As promised in the previous blogpost, here is some finite group theory. (Those of you familiar with the British TV show “Faking It” will appreciate that you don’t have to fool all of the people all of the time, just some of the people at the right moments.)

Some preliminary terminology is useful, since we will need it in later posts. We say that H is (isomorphic to) a proper quotient of a group G if there is a homomorphism from G onto H which has non-trivial kernel. A group G is said to be just non-abelian (or JNA or short) if it is non-abelian, yet every proper quotient is abelian. A little thought shows that this is equivalent to the condition that the derived subgroup [G,G] is contained in each non-trivial normal subgroup of G.

(I don’t remember explicitly hearing about JNA groups in past courses or talks, but about seven years ago in Newcastle I heard a visiting speaker talk about families of groups that were “just-${{\mathcal P}}$” for some property ${{\mathcal P}}$.)

Doing some digging in the literature: in the case where the group G is JNA and [G,G] is abelian, there is a classification or structure theorem available, through work of M. F. Newman in two papers that appear in the Proceedings of the London Mathematical Society (both in volume 10, 1960). Later on, when we resume the story of the central amenability constant, JNA groups will turn up very naturally. However, none of that is needed to follow the proof of the following result:

Theorem 1. Let G be a finite JNA group which has trivial centre and which has a conjugacy class of size 2. Then G contains an involution ${t}$ and an element ${r}$ of odd prime order p, such that G has order 2p and ${rtr=t}$.

Up to isomorphism, the only group satisfying the conclusions of the theorem is the dihedral group of order 2p. (Checking that this group actually is JNA is not too difficult, but I won’t go into the details here.)

Remark. In any finite group with a conjugacy class of size 2, both elements in the conjugacy class have the same centralizer — this point will be reiterated below in a little more detail — and this centralizer has index 2 in the parent group, hence is normal. This is encouraging in our setting, since the JNA condition now gives us extra information.

## A leisurely proof

The purpose of this post is to provide a proof of Theorem 1 that uses only basic facts of finite group theory, such as may be found in a first course that covers notions such as conjugacy classes and normal subgroups.

We recall that the derived subgroup of G, sometimes called the commutator subgroup of G, is the subgroup of G generated by all elements of the form ${xyx^{-1}y^{-1}}$ as ${x,y}$ vary over G. (For those who prefer to think categorically, [G,G] is uniquely determined by the property that it is contained in the kernel of every homomorphism from G to any abelian group, i.e. it is the kernel of the abelianization homomorphism.)

Now in Theorem 1, the hypotheses on G are as follows:

1. Z(G)${=\{e\}}$
2. there exists ${r\in}$ G with exactly one other conjugate, ${\overline{r}\neq r}$
3. If N is a non-trivial normal subgroup of G, then every commutator ${[x,y]:=xyx^{-1}y^{-1}}$ belongs to N.

It is immediate from Condition 2 that any given ${x\in}$ G either centralizes both ${r}$ and ${\overline{r}}$, or else swaps them (this is true for any group action on any 2-point set, of course). More formally:

Lemma 2. Let ${x\in}$ G. Either ${xrx^{-1}=r}$ and ${x\overline{r} x^{-1}=\overline{r}}$, or ${xrx^{-1}=\overline{r}}$ and ${x\overline{r} x^{-1} = r}$.

In particular, this lemma implies that ${r\overline{r} r^{-1}=\overline{r}}$, i.e. ${r}$ and ${\overline{r}}$ commute. Applying the lemma to ${r\overline{r}}$ shows that ${x(r\overline{r})x^{-1}=r\overline{r}}$ for all ${x}$ in G, and so by Condition 1 we must have

$\displaystyle r\overline{r}=e. \ \ \ \ \ (1)$

Let N be the order of ${r}$: this is an integer ${\geq 2}$. If N were even, say N=2m, then ${r^m = r^{-m} = (\overline{r})^m \neq e}$, and applying Lemma 2 we would find that ${r^m\in}$ Z(G), which contradicts Condition 1. Therefore N is odd.

Now we fix ${t}$ in G such that ${\overline{r}=trt^{-1}}$. (It will turn out that ${t}$ has to be an involution, but the proof is somewhat indirect.) By Lemma 2, ${t\overline{r} t^{-1} = r}$, and since ${\overline{r}=r^{-1}}$ (Equation (1)) this implies

$\displaystyle r^2 = tr^{-1}t^{-1}r \in [G,G].$

Because ${r}$ has odd order, this implies that ${r\in [G,G]}$, and so [G,G] contains the subgroup of G generated by ${r}$, which we denote by ${\langle r\rangle}$. On the other hand, observe that since ${\overline{r}\in \langle r\rangle}$, Lemma 2 implies that ${\langle r\rangle}$ is a normal subgroup of G. Therefore, by Condition 3,

$\displaystyle [G,G] = \langle r\rangle. \ \ \ \ \ (2)$

Let H be the centralizer in G of the element ${r}$ (and hence, as remarked above, of the element ${\overline{r}}$). We know that H has index 2 in G (by applying the orbit-stabilizer theorem to the conjugation action of G on ${\{r,\overline{r}\}}$). At this point we could now invoke the general fact that index 2 subgroups of any group are necessarily normal subgroups. To keep things self-contained, we instead use some ad hoc arguments (although this admittedly leaves out the bigger picture which motivates our calculations).

Lemma 3. Let ${x\in}$ G. Then either ${x}$ or ${xt}$ belongs to H.

Proof: If ${x}$ belongs to G but not H, then ${x^{-1}}$ does not lie in H, and so ${x^{-1}rx=\overline{r}}$ by Lemma 2. So ${x^{-1}rx= trt^{-1}}$ and rearranging shows that ${xt}$ centralizes ${r}$. $\Box$

Moreover, by definition H contains ${r}$. Hence it contains [G,G], by Equation (2). The next step in our argument is to show that in fact H=[G,G].

Digression. At this point, if we just wanted to proceed as quickly as possible, we could appeal to Theorem 3.4 of

M. F. Newman, On a class of metabelian groups. Proc. London Math. Soc. (3) 10 1960 354–364. MR0117293 (22 #8074)

Indeed, this point is where I’d originally got stuck on my first attempt to prove the theorem, and I has to resort to Newman’s paper to check that what I was hoping to prove was actually true. Nevertheless, it seems worth giving an ad hoc argument using only “bare-hands techniques”, since we are in a much more specialized setting than that covered by Newman’s theorem. What follows is my own argument, found by some trial and error after I had used Newman’s paper to check I was on the right lines.

### Proof that H=[G,G]

Consider the function θ : H${\rightarrow}$ [G,G] defined by ${\theta(h)=tht^{-1}h^{-1}}$. Since [G,G]=${\langle r\rangle}$ is contained in Z(H), for all ${h,k\in}$ H we have

$\displaystyle t(hk)t^{-1} = (tht^{-1})(tkt^{-1}) = \theta(h)h\theta(k)k=\theta(h)\theta(k)hk.$

Thus

$\displaystyle \theta(hk)=\theta(h)\theta(k)\quad\mbox{for all }h,k\in H; \ \ \ \ \ (3)$

in other words, θ is a homomorphism. Our goal is to show θ is a bijection, which will force H and [G,G] to have the same cardinality. Since we already know that [G,G] is contained in H, we can conclude that [G,G]=H as required.

First, we show θ is surjective. Note that ${\theta(r^{-1}) = tr^{-1}t^{-1}r = r^2}$, so if we let m=(N+1)/2 and k be any integer, induction (or the fact θ is a homomorphism) implies that ${\theta(r^{-mk}) = r^k}$. Since [G,G]=${\langle r\rangle}$, this shows θ(H)=[G,G].

Secondly, we show θ is injective. Let K=${\ker(\theta)}$, which is a normal subgroup of H (being the kernel of a homomorphism). Note that K=${\{h\in H \colon ht=th\}}$. It now follows from Lemma 3 that K is normal as a subgroup of G. Suppose K is not the trivial subgroup; then by Condition 3, K contains [G,G], and in particular ${r}$ belongs to K. But since ${trt^{-1}=\overline{r}=r^{-1}}$ (Equation (1)) we have

$\displaystyle \theta(r) = trt^{-1}r^{-1} = r^{-2} \neq \{e\}$

(since ${r\neq\overline{r}}$) and we get a contradiction. Therefore K is the trivial subgroup, and θ:H${\rightarrow}$[G,G] is indeed injective.

### Continuing the proof of Theorem 1

Let us take stock. We have shown that there exist elements ${r,t\in}$G such that:

• ${trt^{-1}=r^{-1}\neq r}$ (this follows from Equation (1) and the choice of ${t}$);
• ${r}$ has odd order, say N (this follows from the remarks after Equation (1));
• ${\langle r\rangle}$ is an index 2 subgroup in G (this follows from Equation (2) and the result just proved).

This is close to what is needed: it only remains to show that ${t^2=e}$ and that N is prime.

We will show that ${t^2}$ belongs to the centre of G, which combined with Condition 1 forces ${t^2=e}$. To do this, first observe that by Lemma 2, ${t^2}$ centralizes ${r}$, and hence centralizes ${\langle r\rangle}$. But since ${\langle r\rangle}$ is an index 2 subgroup of G and ${t\notin\langle r\rangle}$, every element of G belongs to either ${\langle r\rangle}$ or ${t\langle r\rangle}$, from which it follows that ${t^2}$ centralizes everything in G, as required.

Remark. The argument just given seems the quickest way to do things, given what we have already shown to date. However, it relies on knowing that ${\langle r\rangle}$ has index 2 in G. It may be of some interest to note that one can show ${t^2}$ centralizes G more directly, using only Equations (1) and (2). Thus, let ${x\in}$G}. By Equation (2), there exists some integer k such that

$\displaystyle t^{-1}xtx^{-1} = r^k.$

Since ${trt^{-1}=r^{-1}}$, it follows that

$\displaystyle xtx^{-1}t = t(r^k)t^{-1} = r^{-k}$

and so ${xtx^{-1}t = (t^{-1}xtx^{-1})^{-1} = xt^{-1}x^{-1}t}$. Rearranging gives

$\displaystyle tx^{-1}t^{-1} = t^{-1}xt$

so that ${t^2 x^{-1} = x^{-1}t^2}$. Since ${x}$ is arbitrary in G, ${t^2}$ belongs to the centre of G.

Finally, to finish things off, it suffices to show that N is prime. Let p be a prime factor of N. Then ${\langle r^p\rangle}$ is a proper subgroup of ${\langle r\rangle=[G,G]}$, and it is normal in G by Equation (1), since ${r^{-p}\in \langle r^p\rangle}$. Condition 3 therefore forces ${r^p=e}$, and so N=p. This completes the proof of Theorem 1.

One Comment leave one →
1. 15 June, 2013 6:22 pm

On second thoughts, that talk mentioning “just-P” groups must have been 2004 or 2005, not 2006. I *think* it was given by John Wilson but my memory is very hazy here.