Examples of subgroups where it's nontrivial to show closure under multiplication?
$begingroup$
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.
abstract-algebra group-theory big-list
$endgroup$
|
show 8 more comments
$begingroup$
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.
abstract-algebra group-theory big-list
$endgroup$
$begingroup$
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
|
show 8 more comments
$begingroup$
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.
abstract-algebra group-theory big-list
$endgroup$
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
abstract-algebra group-theory big-list
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
|
show 8 more comments
$begingroup$
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
|
show 8 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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$$begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\a_{21}&a_{22}&a_{23}&a_{24}\0&0&a_{33}&a_{34}\0&0&a_{43}&a_{44}end{bmatrix}.$$
$endgroup$
$begingroup$
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
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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.
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– Ehsaan
31 mins ago
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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.
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– Ehsaan
31 mins ago
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@Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
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– darij grinberg
14 mins ago
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@Ehsaan: I don't think determinants over $mathbb{F}_2$ will help you much...
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– darij grinberg
14 mins ago
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@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbf{Q}$ and argue it is $pm 1$.
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– Ehsaan
10 mins ago
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@darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $mathbf{Q}$ and argue it is $pm 1$.
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– Ehsaan
10 mins ago
add a comment |
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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$$begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\a_{21}&a_{22}&a_{23}&a_{24}\0&0&a_{33}&a_{34}\0&0&a_{43}&a_{44}end{bmatrix}.$$
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$begingroup$
I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise".
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– yamete kudasai
29 mins ago
add a comment |
$begingroup$
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$$begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\a_{21}&a_{22}&a_{23}&a_{24}\0&0&a_{33}&a_{34}\0&0&a_{43}&a_{44}end{bmatrix}.$$
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$begingroup$
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
add a comment |
$begingroup$
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$$begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\a_{21}&a_{22}&a_{23}&a_{24}\0&0&a_{33}&a_{34}\0&0&a_{43}&a_{44}end{bmatrix}.$$
$endgroup$
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$$begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\a_{21}&a_{22}&a_{23}&a_{24}\0&0&a_{33}&a_{34}\0&0&a_{43}&a_{44}end{bmatrix}.$$
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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.
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– YCor
41 mins ago
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The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup.
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– YCor
39 mins ago
1
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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
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– Max
36 mins ago
1
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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).
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– 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.
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– YCor
26 mins ago