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.

  • $\begingroup$ 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
    Apr 27, 2019 at 14:26
  • $\begingroup$ The set of exponentially distorted elements in a simply connected solvable Lie group is a subgroup. $\endgroup$
    – YCor
    Apr 27, 2019 at 14:28
  • 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 $\endgroup$ Apr 27, 2019 at 14:31
  • 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
    Apr 27, 2019 at 14:37
  • 3
    $\begingroup$ 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
    Apr 27, 2019 at 14:41

3 Answers 3


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).

  • 6
    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Apr 27, 2019 at 16:56
  • 1
    $\begingroup$ For those who don't know about John G. Thompson: "more recently" means that he proved it in his PhD thesis 1959. $\endgroup$
    – j.p.
    Apr 28, 2019 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.

  • $\begingroup$ 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
    Apr 27, 2019 at 14:31
  • $\begingroup$ 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
    Apr 27, 2019 at 14:36
  • $\begingroup$ @Ehsaan: I don't think determinants over $\mathbb{F}_2$ will help you much... $\endgroup$ Apr 27, 2019 at 14:53
  • $\begingroup$ @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
    Apr 27, 2019 at 14:57
  • 3
    $\begingroup$ 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. $\endgroup$
    – Ehsaan
    Apr 27, 2019 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$$\begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\0&0&a_{33}&a_{34}\\0&0&a_{43}&a_{44}\end{bmatrix}.$$

  • 5
    $\begingroup$ I think this one follows from the fact that 2x2 upper triangular matrices form a group and that you can operate matrices "blockwise". $\endgroup$ Apr 27, 2019 at 14:38
  • $\begingroup$ Indeed it does, but many Linear Algebra students don't know that. $\endgroup$ Apr 27, 2019 at 16:05

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