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












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
    41 mins ago












  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    39 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
    36 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
    30 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
    26 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
    41 mins ago












  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    39 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
    36 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
    30 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
    26 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 21 mins ago









mrtaurho

6,36071742




6,36071742










asked 1 hour 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
    41 mins ago












  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    39 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
    36 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
    30 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
    26 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
    41 mins ago












  • $begingroup$
    The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
    $endgroup$
    – YCor
    39 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
    36 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
    30 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
    26 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
41 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
41 mins ago














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




$begingroup$
The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
$endgroup$
– YCor
39 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
36 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
36 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
30 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
30 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
26 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
26 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
    36 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
    31 mins ago










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
    $endgroup$
    – darij grinberg
    14 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
    10 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
    29 mins ago














Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3204485%2fexamples-of-subgroups-where-its-nontrivial-to-show-closure-under-multiplication%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























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
    36 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
    31 mins ago










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
    $endgroup$
    – darij grinberg
    14 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
    10 mins ago
















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
    36 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
    31 mins ago










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
    $endgroup$
    – darij grinberg
    14 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
    10 mins ago














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 43 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
    36 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
    31 mins ago










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
    $endgroup$
    – darij grinberg
    14 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
    10 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
    36 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
    31 mins ago










  • $begingroup$
    @Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
    $endgroup$
    – darij grinberg
    14 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
    10 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
36 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
36 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
31 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
31 mins ago












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






$begingroup$
@Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
$endgroup$
– darij grinberg
14 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
10 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
10 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
    29 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
    29 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 52 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
    29 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
    29 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
29 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
29 mins ago




















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3204485%2fexamples-of-subgroups-where-its-nontrivial-to-show-closure-under-multiplication%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Bundesstraße 106

Ida-Boy-Ed-Garten

Verónica Boquete