Examples of subgroups where it's nontrivial to show closure under multiplication?Can a finitely generated group have infinitely many torsion elements?The torsion subset of a non-abelian group is not, in general, a subgroup.If there exists a subgroup contained in all nontrivial subgroups of $G$, then $G$ is quasicyclic.torsion subgroup of a finitely generated nilpotent group is finite.Is a finitely generated torsion group finite in general?Residually finite group with finitely many conjugacy classes of elements of finite orderPart of simple proof of nontrivial center in p-groupWhy $|G:Z(G)|$ is finite in this question?Is the FC-center of a finitely generated group itself finitely generated?Conjugacy classes generating an infinite group

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Examples of subgroups where it's nontrivial to show closure under multiplication?


Can a finitely generated group have infinitely many torsion elements?The torsion subset of a non-abelian group is not, in general, a subgroup.If there exists a subgroup contained in all nontrivial subgroups of $G$, then $G$ is quasicyclic.torsion subgroup of a finitely generated nilpotent group is finite.Is a finitely generated torsion group finite in general?Residually finite group with finitely many conjugacy classes of elements of finite orderPart of simple proof of nontrivial center in p-groupWhy $|G:Z(G)|$ is finite in this question?Is the FC-center of a finitely generated group itself finitely generated?Conjugacy classes generating an infinite group













11












$begingroup$


Usually when a subgroup is declared, it is trivial (or at least straightforward to a sophomore) to prove that it is a subgroup under multiplication. For example:



  • Homomorphic image and preimage of a subgroup

  • Center

  • Intersection of two subgroups

  • Stabilizer of a point in a group action

  • Elements of finite conjugacy class


  • $HN$, where $Hleq G$ and $Ntrianglelefteq G$

I'm looking for interesting theorems where a subset is claimed to be a subgroup, but it is nontrivial to verify. I am looking for some kind of structure-theoretic subgroup that can be defined for any group (or a large class of groups), rather than specific examples that are hard to show closure.




I can think of only one example.



Let $Delta(G)$ be those elements with finite conjugacy class (easily seen to be a subgroup), and let $Delta^+(G)$ be its torsion subset. Note that $g$ has finite conjugacy class in $G$ if and only if $[G:C(g)]<infty$ where $C(g)$ is the centralizer.



$$Delta^+(G):=<infty, [G:C(g)]<infty.$$




Theorem: $Delta^+(G)$ is closed under multiplication.




The proof takes 1-2 pages of nontrivial calculations. If $a,b$ are torsion, it's not true that their product $ab$ is torsion --- but amazingly, it is true if $a,b$ have only finitely many conjugates.



Does anyone have any other examples?




Proof of the Theorem, for those interested. You can see it from the following (nontrivial) lemma.




Lemma (Dietzmann). If $[G:Z(G)]<infty$, then $[G,G]$ is finite.




Modulo the (page long) proof of this, let's see why it implies the theorem.



Let $x,yin Delta^+(G)$ so that $x,y$ have finite conjugacy classes and orders. Clearly $xy$ has a finite conjugacy class, so we just have to show that it has finite order.



Let $N$ be the subgroup generated by all the conjugates of $x$ and $y$, so that $N$ is finitely-generated. Then $N/N'$ is an abelian group generated by finitely many torsion elements, hence finite, so $[N:N']<infty$. It is thus enough to show $N'$ is finite, because then $N$ is finite, and since $xyin N$ this completes the proof.



To show $N'$ is finite we use Dietzmann's Lemma: notice $Z(N)=C_N(x)cap C_N(y)$, and these centralizers have finite index in $N$. Therefore $[N:Z(N)]<infty$ and we apply Dietzmann's Lemma.



This was already a somewhat lengthy and interesting argument, and we haven't even proved Dietzmann's Lemma yet!




Edit: a related question is as follows. Name any functions $varphi:Grightarrow H$ that are homomorphisms, but it is nontrivial to show.










share|cite|improve this question











$endgroup$











  • $begingroup$
    The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:26











  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:28






  • 1




    $begingroup$
    I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
    $endgroup$
    – Max
    Apr 27 at 14:31






  • 1




    $begingroup$
    You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
    $endgroup$
    – YCor
    Apr 27 at 14:37






  • 2




    $begingroup$
    I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:41















11












$begingroup$


Usually when a subgroup is declared, it is trivial (or at least straightforward to a sophomore) to prove that it is a subgroup under multiplication. For example:



  • Homomorphic image and preimage of a subgroup

  • Center

  • Intersection of two subgroups

  • Stabilizer of a point in a group action

  • Elements of finite conjugacy class


  • $HN$, where $Hleq G$ and $Ntrianglelefteq G$

I'm looking for interesting theorems where a subset is claimed to be a subgroup, but it is nontrivial to verify. I am looking for some kind of structure-theoretic subgroup that can be defined for any group (or a large class of groups), rather than specific examples that are hard to show closure.




I can think of only one example.



Let $Delta(G)$ be those elements with finite conjugacy class (easily seen to be a subgroup), and let $Delta^+(G)$ be its torsion subset. Note that $g$ has finite conjugacy class in $G$ if and only if $[G:C(g)]<infty$ where $C(g)$ is the centralizer.



$$Delta^+(G):=<infty, [G:C(g)]<infty.$$




Theorem: $Delta^+(G)$ is closed under multiplication.




The proof takes 1-2 pages of nontrivial calculations. If $a,b$ are torsion, it's not true that their product $ab$ is torsion --- but amazingly, it is true if $a,b$ have only finitely many conjugates.



Does anyone have any other examples?




Proof of the Theorem, for those interested. You can see it from the following (nontrivial) lemma.




Lemma (Dietzmann). If $[G:Z(G)]<infty$, then $[G,G]$ is finite.




Modulo the (page long) proof of this, let's see why it implies the theorem.



Let $x,yin Delta^+(G)$ so that $x,y$ have finite conjugacy classes and orders. Clearly $xy$ has a finite conjugacy class, so we just have to show that it has finite order.



Let $N$ be the subgroup generated by all the conjugates of $x$ and $y$, so that $N$ is finitely-generated. Then $N/N'$ is an abelian group generated by finitely many torsion elements, hence finite, so $[N:N']<infty$. It is thus enough to show $N'$ is finite, because then $N$ is finite, and since $xyin N$ this completes the proof.



To show $N'$ is finite we use Dietzmann's Lemma: notice $Z(N)=C_N(x)cap C_N(y)$, and these centralizers have finite index in $N$. Therefore $[N:Z(N)]<infty$ and we apply Dietzmann's Lemma.



This was already a somewhat lengthy and interesting argument, and we haven't even proved Dietzmann's Lemma yet!




Edit: a related question is as follows. Name any functions $varphi:Grightarrow H$ that are homomorphisms, but it is nontrivial to show.










share|cite|improve this question











$endgroup$











  • $begingroup$
    The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:26











  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:28






  • 1




    $begingroup$
    I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
    $endgroup$
    – Max
    Apr 27 at 14:31






  • 1




    $begingroup$
    You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
    $endgroup$
    – YCor
    Apr 27 at 14:37






  • 2




    $begingroup$
    I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:41













11












11








11


4



$begingroup$


Usually when a subgroup is declared, it is trivial (or at least straightforward to a sophomore) to prove that it is a subgroup under multiplication. For example:



  • Homomorphic image and preimage of a subgroup

  • Center

  • Intersection of two subgroups

  • Stabilizer of a point in a group action

  • Elements of finite conjugacy class


  • $HN$, where $Hleq G$ and $Ntrianglelefteq G$

I'm looking for interesting theorems where a subset is claimed to be a subgroup, but it is nontrivial to verify. I am looking for some kind of structure-theoretic subgroup that can be defined for any group (or a large class of groups), rather than specific examples that are hard to show closure.




I can think of only one example.



Let $Delta(G)$ be those elements with finite conjugacy class (easily seen to be a subgroup), and let $Delta^+(G)$ be its torsion subset. Note that $g$ has finite conjugacy class in $G$ if and only if $[G:C(g)]<infty$ where $C(g)$ is the centralizer.



$$Delta^+(G):=<infty, [G:C(g)]<infty.$$




Theorem: $Delta^+(G)$ is closed under multiplication.




The proof takes 1-2 pages of nontrivial calculations. If $a,b$ are torsion, it's not true that their product $ab$ is torsion --- but amazingly, it is true if $a,b$ have only finitely many conjugates.



Does anyone have any other examples?




Proof of the Theorem, for those interested. You can see it from the following (nontrivial) lemma.




Lemma (Dietzmann). If $[G:Z(G)]<infty$, then $[G,G]$ is finite.




Modulo the (page long) proof of this, let's see why it implies the theorem.



Let $x,yin Delta^+(G)$ so that $x,y$ have finite conjugacy classes and orders. Clearly $xy$ has a finite conjugacy class, so we just have to show that it has finite order.



Let $N$ be the subgroup generated by all the conjugates of $x$ and $y$, so that $N$ is finitely-generated. Then $N/N'$ is an abelian group generated by finitely many torsion elements, hence finite, so $[N:N']<infty$. It is thus enough to show $N'$ is finite, because then $N$ is finite, and since $xyin N$ this completes the proof.



To show $N'$ is finite we use Dietzmann's Lemma: notice $Z(N)=C_N(x)cap C_N(y)$, and these centralizers have finite index in $N$. Therefore $[N:Z(N)]<infty$ and we apply Dietzmann's Lemma.



This was already a somewhat lengthy and interesting argument, and we haven't even proved Dietzmann's Lemma yet!




Edit: a related question is as follows. Name any functions $varphi:Grightarrow H$ that are homomorphisms, but it is nontrivial to show.










share|cite|improve this question











$endgroup$




Usually when a subgroup is declared, it is trivial (or at least straightforward to a sophomore) to prove that it is a subgroup under multiplication. For example:



  • Homomorphic image and preimage of a subgroup

  • Center

  • Intersection of two subgroups

  • Stabilizer of a point in a group action

  • Elements of finite conjugacy class


  • $HN$, where $Hleq G$ and $Ntrianglelefteq G$

I'm looking for interesting theorems where a subset is claimed to be a subgroup, but it is nontrivial to verify. I am looking for some kind of structure-theoretic subgroup that can be defined for any group (or a large class of groups), rather than specific examples that are hard to show closure.




I can think of only one example.



Let $Delta(G)$ be those elements with finite conjugacy class (easily seen to be a subgroup), and let $Delta^+(G)$ be its torsion subset. Note that $g$ has finite conjugacy class in $G$ if and only if $[G:C(g)]<infty$ where $C(g)$ is the centralizer.



$$Delta^+(G):=<infty, [G:C(g)]<infty.$$




Theorem: $Delta^+(G)$ is closed under multiplication.




The proof takes 1-2 pages of nontrivial calculations. If $a,b$ are torsion, it's not true that their product $ab$ is torsion --- but amazingly, it is true if $a,b$ have only finitely many conjugates.



Does anyone have any other examples?




Proof of the Theorem, for those interested. You can see it from the following (nontrivial) lemma.




Lemma (Dietzmann). If $[G:Z(G)]<infty$, then $[G,G]$ is finite.




Modulo the (page long) proof of this, let's see why it implies the theorem.



Let $x,yin Delta^+(G)$ so that $x,y$ have finite conjugacy classes and orders. Clearly $xy$ has a finite conjugacy class, so we just have to show that it has finite order.



Let $N$ be the subgroup generated by all the conjugates of $x$ and $y$, so that $N$ is finitely-generated. Then $N/N'$ is an abelian group generated by finitely many torsion elements, hence finite, so $[N:N']<infty$. It is thus enough to show $N'$ is finite, because then $N$ is finite, and since $xyin N$ this completes the proof.



To show $N'$ is finite we use Dietzmann's Lemma: notice $Z(N)=C_N(x)cap C_N(y)$, and these centralizers have finite index in $N$. Therefore $[N:Z(N)]<infty$ and we apply Dietzmann's Lemma.



This was already a somewhat lengthy and interesting argument, and we haven't even proved Dietzmann's Lemma yet!




Edit: a related question is as follows. Name any functions $varphi:Grightarrow H$ that are homomorphisms, but it is nontrivial to show.







abstract-algebra group-theory examples-counterexamples big-list






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 28 at 10:45









Martin Sleziak

45.2k11123278




45.2k11123278










asked Apr 27 at 14:07









EhsaanEhsaan

1,381515




1,381515











  • $begingroup$
    The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:26











  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:28






  • 1




    $begingroup$
    I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
    $endgroup$
    – Max
    Apr 27 at 14:31






  • 1




    $begingroup$
    You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
    $endgroup$
    – YCor
    Apr 27 at 14:37






  • 2




    $begingroup$
    I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:41
















  • $begingroup$
    The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:26











  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:28






  • 1




    $begingroup$
    I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
    $endgroup$
    – Max
    Apr 27 at 14:31






  • 1




    $begingroup$
    You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
    $endgroup$
    – YCor
    Apr 27 at 14:37






  • 2




    $begingroup$
    I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
    $endgroup$
    – YCor
    Apr 27 at 14:41















$begingroup$
The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:26





$begingroup$
The set of torsion elements in a nilpotent group is a subgroup. And the set of elements of order some power of a prime $p$, in a nilpotent group, is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:26













$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:28




$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:28




1




1




$begingroup$
I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
$endgroup$
– Max
Apr 27 at 14:31




$begingroup$
I don't remember the precise statement, but there is a theorem that looks like "let the finite group $G$ act transitively on the finite set $X$ such that every element of $G$ has at most one fixed point. Then the set of elements of $G$ that have no fixed point + $e$ is a subgroup of $G$" - if I remember correctly, the proof uses representation theory
$endgroup$
– Max
Apr 27 at 14:31




1




1




$begingroup$
You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
$endgroup$
– YCor
Apr 27 at 14:37




$begingroup$
You don't need any use of structure theorem for f.g. abelian groups (clearly every f.g. abelian group generated by torsion elements is finite).
$endgroup$
– YCor
Apr 27 at 14:37




2




2




$begingroup$
I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:41




$begingroup$
I don't know if it qualifies, but it is a very difficult and recent theorem (positive solution to Ore's conjecture) that the set of commutators in every finite simple group is a subgroup.
$endgroup$
– YCor
Apr 27 at 14:41










3 Answers
3






active

oldest

votes


















8












$begingroup$

A particularly nice example is the following : suppose the finite group $G$ acts on the finite set $X$ in such a way that every nontrivial element of $G$ has at most one fixed point. Let $S$ be the set of elements of $G$ that have no fixed points. Then $H=Scup 1$ is a subgroup of $G$.



I believe the only known proofs are representation-theoretic (or at least that was the case at first).






share|cite|improve this answer









$endgroup$








  • 5




    $begingroup$
    Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
    $endgroup$
    – Derek Holt
    Apr 27 at 16:56






  • 1




    $begingroup$
    For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
    $endgroup$
    – j.p.
    Apr 28 at 8:19


















4












$begingroup$

Consider the symmetric group $S_n$ on $ngeq 2$ letters. The alternating group $A_n$ is the subgroup of $S_n$ given by all even permutations of $S_n$.



One proof uses the signum or sign function $s:S_nrightarrowpm 1$ which assigns to a permutation $pi$, $+1$ if $pi$ is even, and $-1$ if $pi$ is odd.
It can be shown that the sign function is a homomorphism, i.e., $s(pisigma) = s(pi)cdot s(sigma)$.



It follows that the product of two even permutations is even and so the multiplication in $A_n$ is well-defined.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:31










  • $begingroup$
    When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:36










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
    $endgroup$
    – darij grinberg
    Apr 27 at 14:53











  • $begingroup$
    @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:57






  • 2




    $begingroup$
    An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
    $endgroup$
    – Ehsaan
    Apr 27 at 16:24


















3












$begingroup$

Let $G=GL(4,k)$ (the group of all $4times4$ invertible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$






share|cite|improve this answer











$endgroup$








  • 5




    $begingroup$
    I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
    $endgroup$
    – yamete kudasai
    Apr 27 at 14:38











  • $begingroup$
    Indeed it does, but many Linear Algebra students don't know that.
    $endgroup$
    – José Carlos Santos
    Apr 27 at 16:05












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









8












$begingroup$

A particularly nice example is the following : suppose the finite group $G$ acts on the finite set $X$ in such a way that every nontrivial element of $G$ has at most one fixed point. Let $S$ be the set of elements of $G$ that have no fixed points. Then $H=Scup 1$ is a subgroup of $G$.



I believe the only known proofs are representation-theoretic (or at least that was the case at first).






share|cite|improve this answer









$endgroup$








  • 5




    $begingroup$
    Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
    $endgroup$
    – Derek Holt
    Apr 27 at 16:56






  • 1




    $begingroup$
    For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
    $endgroup$
    – j.p.
    Apr 28 at 8:19















8












$begingroup$

A particularly nice example is the following : suppose the finite group $G$ acts on the finite set $X$ in such a way that every nontrivial element of $G$ has at most one fixed point. Let $S$ be the set of elements of $G$ that have no fixed points. Then $H=Scup 1$ is a subgroup of $G$.



I believe the only known proofs are representation-theoretic (or at least that was the case at first).






share|cite|improve this answer









$endgroup$








  • 5




    $begingroup$
    Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
    $endgroup$
    – Derek Holt
    Apr 27 at 16:56






  • 1




    $begingroup$
    For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
    $endgroup$
    – j.p.
    Apr 28 at 8:19













8












8








8





$begingroup$

A particularly nice example is the following : suppose the finite group $G$ acts on the finite set $X$ in such a way that every nontrivial element of $G$ has at most one fixed point. Let $S$ be the set of elements of $G$ that have no fixed points. Then $H=Scup 1$ is a subgroup of $G$.



I believe the only known proofs are representation-theoretic (or at least that was the case at first).






share|cite|improve this answer









$endgroup$



A particularly nice example is the following : suppose the finite group $G$ acts on the finite set $X$ in such a way that every nontrivial element of $G$ has at most one fixed point. Let $S$ be the set of elements of $G$ that have no fixed points. Then $H=Scup 1$ is a subgroup of $G$.



I believe the only known proofs are representation-theoretic (or at least that was the case at first).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 27 at 15:58









MaxMax

17.5k11144




17.5k11144







  • 5




    $begingroup$
    Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
    $endgroup$
    – Derek Holt
    Apr 27 at 16:56






  • 1




    $begingroup$
    For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
    $endgroup$
    – j.p.
    Apr 28 at 8:19












  • 5




    $begingroup$
    Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
    $endgroup$
    – Derek Holt
    Apr 27 at 16:56






  • 1




    $begingroup$
    For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
    $endgroup$
    – j.p.
    Apr 28 at 8:19







5




5




$begingroup$
Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
$endgroup$
– Derek Holt
Apr 27 at 16:56




$begingroup$
Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent.
$endgroup$
– Derek Holt
Apr 27 at 16:56




1




1




$begingroup$
For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
$endgroup$
– j.p.
Apr 28 at 8:19




$begingroup$
For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959.
$endgroup$
– j.p.
Apr 28 at 8:19











4












$begingroup$

Consider the symmetric group $S_n$ on $ngeq 2$ letters. The alternating group $A_n$ is the subgroup of $S_n$ given by all even permutations of $S_n$.



One proof uses the signum or sign function $s:S_nrightarrowpm 1$ which assigns to a permutation $pi$, $+1$ if $pi$ is even, and $-1$ if $pi$ is odd.
It can be shown that the sign function is a homomorphism, i.e., $s(pisigma) = s(pi)cdot s(sigma)$.



It follows that the product of two even permutations is even and so the multiplication in $A_n$ is well-defined.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:31










  • $begingroup$
    When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:36










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
    $endgroup$
    – darij grinberg
    Apr 27 at 14:53











  • $begingroup$
    @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:57






  • 2




    $begingroup$
    An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
    $endgroup$
    – Ehsaan
    Apr 27 at 16:24















4












$begingroup$

Consider the symmetric group $S_n$ on $ngeq 2$ letters. The alternating group $A_n$ is the subgroup of $S_n$ given by all even permutations of $S_n$.



One proof uses the signum or sign function $s:S_nrightarrowpm 1$ which assigns to a permutation $pi$, $+1$ if $pi$ is even, and $-1$ if $pi$ is odd.
It can be shown that the sign function is a homomorphism, i.e., $s(pisigma) = s(pi)cdot s(sigma)$.



It follows that the product of two even permutations is even and so the multiplication in $A_n$ is well-defined.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:31










  • $begingroup$
    When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:36










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
    $endgroup$
    – darij grinberg
    Apr 27 at 14:53











  • $begingroup$
    @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:57






  • 2




    $begingroup$
    An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
    $endgroup$
    – Ehsaan
    Apr 27 at 16:24













4












4








4





$begingroup$

Consider the symmetric group $S_n$ on $ngeq 2$ letters. The alternating group $A_n$ is the subgroup of $S_n$ given by all even permutations of $S_n$.



One proof uses the signum or sign function $s:S_nrightarrowpm 1$ which assigns to a permutation $pi$, $+1$ if $pi$ is even, and $-1$ if $pi$ is odd.
It can be shown that the sign function is a homomorphism, i.e., $s(pisigma) = s(pi)cdot s(sigma)$.



It follows that the product of two even permutations is even and so the multiplication in $A_n$ is well-defined.






share|cite|improve this answer









$endgroup$



Consider the symmetric group $S_n$ on $ngeq 2$ letters. The alternating group $A_n$ is the subgroup of $S_n$ given by all even permutations of $S_n$.



One proof uses the signum or sign function $s:S_nrightarrowpm 1$ which assigns to a permutation $pi$, $+1$ if $pi$ is even, and $-1$ if $pi$ is odd.
It can be shown that the sign function is a homomorphism, i.e., $s(pisigma) = s(pi)cdot s(sigma)$.



It follows that the product of two even permutations is even and so the multiplication in $A_n$ is well-defined.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Apr 27 at 14:24









WuestenfuxWuestenfux

6,1761513




6,1761513











  • $begingroup$
    Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:31










  • $begingroup$
    When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:36










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
    $endgroup$
    – darij grinberg
    Apr 27 at 14:53











  • $begingroup$
    @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:57






  • 2




    $begingroup$
    An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
    $endgroup$
    – Ehsaan
    Apr 27 at 16:24
















  • $begingroup$
    Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:31










  • $begingroup$
    When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:36










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
    $endgroup$
    – darij grinberg
    Apr 27 at 14:53











  • $begingroup$
    @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
    $endgroup$
    – Ehsaan
    Apr 27 at 14:57






  • 2




    $begingroup$
    An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
    $endgroup$
    – Ehsaan
    Apr 27 at 16:24















$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
$endgroup$
– Ehsaan
Apr 27 at 14:31




$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbfF_2$) to construct the sign function, which makes it "easy" to show that it's a homomorphism and its kernel is a subgroup of index $2$, hence normal.
$endgroup$
– Ehsaan
Apr 27 at 14:31












$begingroup$
When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
$endgroup$
– Ehsaan
Apr 27 at 14:36




$begingroup$
When I first took group theory, the instructor painstakingly showed that if a permutation is a product of transpositions, then the number of such transpositions is unique modulo $2$. I'm not sure if using determinants properly circumvents this technical argument, or sweeps it under the rug.
$endgroup$
– Ehsaan
Apr 27 at 14:36












$begingroup$
@Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
$endgroup$
– darij grinberg
Apr 27 at 14:53





$begingroup$
@Ehsaan: I don't think determinants over $mathbbF_2$ will help you much...
$endgroup$
– darij grinberg
Apr 27 at 14:53













$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
Apr 27 at 14:57




$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbfQ$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
Apr 27 at 14:57




2




2




$begingroup$
An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
$endgroup$
– Ehsaan
Apr 27 at 16:24




$begingroup$
An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd.
$endgroup$
– Ehsaan
Apr 27 at 16:24











3












$begingroup$

Let $G=GL(4,k)$ (the group of all $4times4$ invertible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$






share|cite|improve this answer











$endgroup$








  • 5




    $begingroup$
    I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
    $endgroup$
    – yamete kudasai
    Apr 27 at 14:38











  • $begingroup$
    Indeed it does, but many Linear Algebra students don't know that.
    $endgroup$
    – José Carlos Santos
    Apr 27 at 16:05
















3












$begingroup$

Let $G=GL(4,k)$ (the group of all $4times4$ invertible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$






share|cite|improve this answer











$endgroup$








  • 5




    $begingroup$
    I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
    $endgroup$
    – yamete kudasai
    Apr 27 at 14:38











  • $begingroup$
    Indeed it does, but many Linear Algebra students don't know that.
    $endgroup$
    – José Carlos Santos
    Apr 27 at 16:05














3












3








3





$begingroup$

Let $G=GL(4,k)$ (the group of all $4times4$ invertible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$






share|cite|improve this answer











$endgroup$



Let $G=GL(4,k)$ (the group of all $4times4$ invertible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 28 at 7:07

























answered Apr 27 at 14:15









José Carlos SantosJosé Carlos Santos

181k24141256




181k24141256







  • 5




    $begingroup$
    I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
    $endgroup$
    – yamete kudasai
    Apr 27 at 14:38











  • $begingroup$
    Indeed it does, but many Linear Algebra students don't know that.
    $endgroup$
    – José Carlos Santos
    Apr 27 at 16:05













  • 5




    $begingroup$
    I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
    $endgroup$
    – yamete kudasai
    Apr 27 at 14:38











  • $begingroup$
    Indeed it does, but many Linear Algebra students don't know that.
    $endgroup$
    – José Carlos Santos
    Apr 27 at 16:05








5




5




$begingroup$
I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
$endgroup$
– yamete kudasai
Apr 27 at 14:38





$begingroup$
I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
$endgroup$
– yamete kudasai
Apr 27 at 14:38













$begingroup$
Indeed it does, but many Linear Algebra students don't know that.
$endgroup$
– José Carlos Santos
Apr 27 at 16:05





$begingroup$
Indeed it does, but many Linear Algebra students don't know that.
$endgroup$
– José Carlos Santos
Apr 27 at 16:05


















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