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

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.

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?

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!

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

• 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. – YCor Apr 27 '19 at 14:26
• The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup. – YCor Apr 27 '19 at 14:28
• 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 – Maxime Ramzi Apr 27 '19 at 14:31
• 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). – YCor Apr 27 '19 at 14:37
• 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. – YCor Apr 27 '19 at 14:41

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).

• Such groups are called Frobenius groups, and the subgroup in question is the Frobenius kernel. It was proved to be a subgroup (using representation theory) by Ftrobenius in 1901. More recently, Thompson proved that this subgroup is nilpotent. – Derek Holt Apr 27 '19 at 16:56
• For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959. – j.p. Apr 28 '19 at 8:19

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.

• 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. – Ehsaan Apr 27 '19 at 14:31
• 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. – Ehsaan Apr 27 '19 at 14:36
• @Ehsaan: I don't think determinants over $\mathbb{F}_2$ will help you much... – darij grinberg Apr 27 '19 at 14:53
• @darijgrinberg Sorry you're right, don't know what I was thinking. Do it over $\mathbf{Q}$ and argue it is $\pm 1$. – Ehsaan Apr 27 '19 at 14:57
• An even (resp. odd) permutation is a product of an even (resp. odd) number of transpositions. The hard part is showing that no permutation is both even and odd. – Ehsaan Apr 27 '19 at 16:24

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$$\begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\0&0&a_{33}&a_{34}\\0&0&a_{43}&a_{44}\end{bmatrix}.$$

• I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise". – nowhere dense Apr 27 '19 at 14:38
• Indeed it does, but many Linear Algebra students don't know that. – José Carlos Santos Apr 27 '19 at 16:05