Mdtryhht

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.

$$Delta^+(G):=langle grangle.$$

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

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

Does anyone have any other examples?

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

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

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

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

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

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

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

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

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

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

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

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$$beginbmatrixa_11&a_12&a_13&a_14\a_21&a_22&a_23&a_24\0&0&a_33&a_34\0&0&a_43&a_44endbmatrix.$$

Generalizing another answer, for any Coxeter system $(W, S) $ there is an alternating subgroup defined as the kernel of the homomorphism $Wto-1,1$ sending the elements of the generating set $S$ to $-1$. Most Coxeter groups don't have a representation that is as nice as the symmetric group, so to prove this is a homomorphism we basically have to construct the root system out of the reflections.