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

Question

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 $H\leq G$ and $N\trianglelefteq G$

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.

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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.

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

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?

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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 $x,y\in \Delta^+(G)$ 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 $xy\in N$ 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!

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

Answer 1

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

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

Answer 2

Consider the symmetric group $S_n$ on $n\geq 2$ 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 $s:S_n\rightarrow\{\pm 1\}$ 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., $s(\pi\sigma) = s(\pi)\cdot s(\sigma)$.

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

Answer 3

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