Examples of subgroups where it's nontrivial to show closure under multiplication?Can a finitely generated...

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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.Is a finitely generated torsion group finite in general?Prove that every finite group G contains a (unique) soluble normal subgroup N such that G/N has no nontrivial abelian normal subgroups.Why $|G:Z(G)|$ is finite in this question?Conjugacy classes generating an infinite groupDoes a virtually abelian group contain a normal finite index abelian subgroup?Torsion subgroup of a finitely-generated abelian group is finite?If $G=Atimes B$, where $A$- torsion group, $B$-free abelian group, how to show that $A=G_{tor}$













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.



$$Delta^+(G):={gin G : |langle grangle|<infty, [G:C(g)]<infty}.$$




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 a 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
    30 mins ago












  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    29 mins ago






  • 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
    26 mins ago






  • 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
    19 mins ago






  • 1




    $begingroup$
    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
    15 mins ago
















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.



$$Delta^+(G):={gin G : |langle grangle|<infty, [G:C(g)]<infty}.$$




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 a 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
    30 mins ago












  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    29 mins ago






  • 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
    26 mins ago






  • 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
    19 mins ago






  • 1




    $begingroup$
    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
    15 mins ago














4












4








4


1



$begingroup$


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.



$$Delta^+(G):={gin G : |langle grangle|<infty, [G:C(g)]<infty}.$$




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 a 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.



$$Delta^+(G):={gin G : |langle grangle|<infty, [G:C(g)]<infty}.$$




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 a 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 big-list






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 11 mins ago









mrtaurho

6,36071742




6,36071742










asked 49 mins ago









EhsaanEhsaan

1,138514




1,138514












  • $begingroup$
    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
    30 mins ago












  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    29 mins ago






  • 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
    26 mins ago






  • 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
    19 mins ago






  • 1




    $begingroup$
    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
    15 mins ago


















  • $begingroup$
    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
    30 mins ago












  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    29 mins ago






  • 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
    26 mins ago






  • 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
    19 mins ago






  • 1




    $begingroup$
    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
    15 mins ago
















$begingroup$
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
30 mins ago






$begingroup$
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
30 mins ago














$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
29 mins ago




$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
29 mins ago




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
26 mins ago




$begingroup$
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
26 mins ago




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
19 mins ago




$begingroup$
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
19 mins ago




1




1




$begingroup$
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
15 mins ago




$begingroup$
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
15 mins ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

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 $s:S_nrightarrow{pm 1}$ 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 $mathbf{F}_2$) 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
    25 mins ago










  • $begingroup$
    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
    21 mins ago










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
    $endgroup$
    – darij grinberg
    4 mins ago












  • $begingroup$
    @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbf{Q}$ and argue it is $pm 1$.
    $endgroup$
    – Ehsaan
    just now



















1












$begingroup$

Let $G=GL(4,k)$ (the group of all $4times4$ inferible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\a_{21}&a_{22}&a_{23}&a_{24}\0&0&a_{33}&a_{34}\0&0&a_{43}&a_{44}end{bmatrix}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    18 mins ago














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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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 $s:S_nrightarrow{pm 1}$ 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 $mathbf{F}_2$) 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
    25 mins ago










  • $begingroup$
    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
    21 mins ago










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
    $endgroup$
    – darij grinberg
    4 mins ago












  • $begingroup$
    @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbf{Q}$ and argue it is $pm 1$.
    $endgroup$
    – Ehsaan
    just now
















2












$begingroup$

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 $s:S_nrightarrow{pm 1}$ 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 $mathbf{F}_2$) 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
    25 mins ago










  • $begingroup$
    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
    21 mins ago










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
    $endgroup$
    – darij grinberg
    4 mins ago












  • $begingroup$
    @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbf{Q}$ and argue it is $pm 1$.
    $endgroup$
    – Ehsaan
    just now














2












2








2





$begingroup$

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 $s:S_nrightarrow{pm 1}$ 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 $s:S_nrightarrow{pm 1}$ 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 33 mins ago









WuestenfuxWuestenfux

5,7501513




5,7501513












  • $begingroup$
    Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbf{F}_2$) 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
    25 mins ago










  • $begingroup$
    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
    21 mins ago










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
    $endgroup$
    – darij grinberg
    4 mins ago












  • $begingroup$
    @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbf{Q}$ and argue it is $pm 1$.
    $endgroup$
    – Ehsaan
    just now


















  • $begingroup$
    Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbf{F}_2$) 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
    25 mins ago










  • $begingroup$
    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
    21 mins ago










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
    $endgroup$
    – darij grinberg
    4 mins ago












  • $begingroup$
    @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbf{Q}$ and argue it is $pm 1$.
    $endgroup$
    – Ehsaan
    just now
















$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbf{F}_2$) 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
25 mins ago




$begingroup$
Ah yeah, this is a tricky one too. You can also use determinant of the associated permutation matrix (considered over $mathbf{F}_2$) 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
25 mins ago












$begingroup$
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
21 mins ago




$begingroup$
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
21 mins ago












$begingroup$
@Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
$endgroup$
– darij grinberg
4 mins ago






$begingroup$
@Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
$endgroup$
– darij grinberg
4 mins ago














$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbf{Q}$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
just now




$begingroup$
@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbf{Q}$ and argue it is $pm 1$.
$endgroup$
– Ehsaan
just now











1












$begingroup$

Let $G=GL(4,k)$ (the group of all $4times4$ inferible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\a_{21}&a_{22}&a_{23}&a_{24}\0&0&a_{33}&a_{34}\0&0&a_{43}&a_{44}end{bmatrix}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    18 mins ago


















1












$begingroup$

Let $G=GL(4,k)$ (the group of all $4times4$ inferible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\a_{21}&a_{22}&a_{23}&a_{24}\0&0&a_{33}&a_{34}\0&0&a_{43}&a_{44}end{bmatrix}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    18 mins ago
















1












1








1





$begingroup$

Let $G=GL(4,k)$ (the group of all $4times4$ inferible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\a_{21}&a_{22}&a_{23}&a_{24}\0&0&a_{33}&a_{34}\0&0&a_{43}&a_{44}end{bmatrix}.$$






share|cite|improve this answer









$endgroup$



Let $G=GL(4,k)$ (the group of all $4times4$ inferible matrices with entries in a field $k$) and let $N$ be the subgroup of those matrices $Min GL(4,k)$ of the form$$begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\a_{21}&a_{22}&a_{23}&a_{24}\0&0&a_{33}&a_{34}\0&0&a_{43}&a_{44}end{bmatrix}.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 41 mins ago









José Carlos SantosJosé Carlos Santos

178k24139251




178k24139251












  • $begingroup$
    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
    18 mins ago




















  • $begingroup$
    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
    18 mins ago


















$begingroup$
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
18 mins ago






$begingroup$
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
18 mins ago




















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