# The number of conjugacy classes and the order of the group

In my response to the OP in Embedding $S_3$ into $Aut(F_2)$ (I continue notation from my response in that thread), I indicated the possibility that lifting elements from $GL_{2}(\mathbb{Z}/(p))$ to $Aut(B(2,p))$ could extend the ability of character-free methods in proving congruences relating the order of a group and its number of conjugacy classes. I can now report a small bit of concrete progress in this direction. Specifically, what I can prove is:

Theorem. If $G$ is a finite group of exponent 3, then $|G| \equiv c(G) \mod{16}$.

Proof. We regard the Burnside group $B(2,3) = < x,y|(*)^{3}=1 >$ (here $*$ ranges over every word in the generators) as $< x,y>$ and specify elements of $Aut(B(2,3))$ by their action on $x$ and $y$. Specifically, consider the following elements of $Aut(B(2,3))$:

$\alpha: (x,y) \to (xy,x^{-1}y)$
$\beta: (x,y) \to (x,y^{-1})$

By checking that $\beta^{2} = 1$ and $\beta\alpha\beta = \alpha^{3}$, we see that $<\alpha,\beta>$ is a quotient of the semidihedral group of order $16$. By checking that $\alpha^{4}$ sends $(x,y)$ to $(x^{-1},xy^{-1}x^{-1})$ so that $\alpha^{4} \neq 1$, we see that $<\alpha,\beta>$ is semidihedral of order $16$.

Now let $<\alpha,\beta>$ act on non-commuting pairs of elements of our group $G$ of exponent 3. We wish to show that every non-identity element of $<\alpha,\beta>$ has no fixed points in its action on the set of non-commuting pairs of elements of $G$. Every non-identity element of $<\alpha,\beta>$ has a power equal to $\alpha^{4}$ or is conjugate to $\beta$, so it suffices to check this property for just $\alpha^{4}$ and $\beta$.

If $(x,y)$ is a fixed point of $\alpha$, then $x = x^{-1}$ so $x^{2} = 1$ and, since $x^{3} = 1$, $x = 1$ and $(x,y)$ is not a non-commuting pair.
If $(x,y)$ is a fixed point of $\beta$, then $y = y^{-1}$ so $y^{2} = 1$ and, since $y^{3} = 1$, $y = 1$ and $(x,y)$ is not a non-commuting pair.

Therefore every orbit for the action of $<\alpha,\beta>$ on the set of non-commuting pairs of elements of $G$ has exactly 16 elements and we conclude that 16 divides $|G|(|G| - c(G))$ . Since $|G|$ is odd, we conclude that $|G| \equiv c(G) \mod{16}$.

I do not know how to give a character-free proof of this congruence (either for general $p$-groups or just those of exponent $p$) for any prime $p$ with $p \equiv 3 \mod{4}$ and $p > 3$. Complicating hopes of extending this is the fact that $B(2,n)$ is not known to be finite for any larger odd value of $n$.

(i) I'm probably just not thinking clearly enough right now, but how does one use this to prove the congruence when $G$ is an arbitrary 3-group?
(ii) Is it hopeless to expect to extend this to larger primes congruent to 3 mod 4?

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I believe $\alpha$ is not an automorphism! – Steve D Oct 8 '11 at 19:20
Thanks! I mistranscribed the formula for $\alpha$. Will edit now. – DavidLHarden Oct 8 '11 at 23:24
When I first read this I missed the "character free" part, so I thought it's pretty simple. I only realized this after clicking on the linked question. :) (This explains why there is a deleted answer of mine pointing to the obvious proof.) – Gjergji Zaimi Oct 22 '11 at 1:34

Burnside's classical formula $|G| \equiv c(G) \text{ mod } 16$ has been generalized by Hirsch in the paper

"On a theorem of Burnside, Quart. J. Math. Oxford 1 (1950), 97-99".

By elementary methods (without using characters or other techniques from representation theory) he shows in the main theorem:

Let $p_1,...,p_k$ be the prime-divisors of $|G|$ and let $d$ be the greatest common divisor of the numbers $p_i^2-1$ $(i=1,...,k)$. Then $$\begin{array}{ll} |G| \equiv c(G) \text{ mod } 2d, & \text{ if } |G| \text{ is odd} \newline |G| \equiv c(G) \text{ mod } 3, & \text{ if } |G| \text{ is even and } 3, |G| \text{ are coprime } \end{array}$$

This also generalizes Geoff Robinson's note for $p$-groups with $p \equiv \pm 1(8)$.

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It is not hopeless to extend this argument so that it applies to all finite groups of odd prime power order. Let $p$ be an odd prime, and let $K$ be a $p$-group. The congruence $|K| \equiv c(K) \mod{16}$ can be proven as a consequence of the stronger (when $p > 3$) congruence $|K| \equiv c(K) \mod{(p^{2}-1)(p-1)}$. This congruence is the best possible in the sense that there are $p$-groups (nonabelian groups of order $p^{3}$) for which $|K| - c(K) = (p^{2}-1)(p-1)$.

Let the exponent of $K$ be $p^{k}$, and let $L$ be the nilpotence class of $K$. It is difficult to work with $B(2,p^{k})$ or $Aut(B(2,p^{k}))$ when $p^{k} > 3$ and neither of those groups is known to be finite, or, worse, when $p^{k} \geq 673$ and they are known to be infinite.
Instead, we build from $K$ the group

$P = < x,y| w^{p^{k}}=1, [w_{1}, \ldots , w_{L+1}]=1 >$,
where $w, w_{1}, \ldots , w_{L+1}$ range independently over every possible word in the generators.

Claim. $P$ is finite.
Proof of Claim. We regard $k$ as fixed and proceed by induction on $L$. When $L = 1$, $P$ is abelian and generated by two elements of order dividing $p^{k}$ so $|P| \leq p^{2k}$ and the base case is proved.
If $L > 1$, it suffices to prove that the group $P_{L}$ in the descending central series of $P$ is finitely generated, since it is abelian and all its elements have finite order dividing $p^{k}$. Then every $(L-1)$-fold commutator $[w_{1}, \ldots , w_{L}]$ is central so $P/P_{L}$ is a quotient of the group which has the same definition as $P$, except for the replacement of $L$ by $L-1$. Since that group is finite by induction, $P/P_{L}$ is finite and the finiteness of $P_{L}$ implies the finiteness of $P$. To prove this, we use the

Commutator Lemma. If $G = < g_{1}, \ldots , g_{m} >$ and $H = < h_{1}, \ldots , h_{r} > \leq G$ for some positive integers $m$ and $r$, then the commutator subgroup $[G,H]$ has a generating set consisting of iterated commutators $[e_{1}, \ldots , e_{C}]$, where each $e_{i} \in$ {$g_{1}^{\pm 1}, \ldots , g_{m}^{\pm 1}, h_{1}^{\pm 1}, \ldots , h_{r}^{\pm 1}$}.

Proof of Commutator Lemma.
Recall the commutator identities $[t, uv] = [t,v][t,u][t,u,v]$ and $[tu, v] = [t,v][t,v,u][u,v]$.
An arbitrary element of $[G,H]$ is a product of commutators of the form $[g,h]$, where $g = \gamma_{1} \ldots \gamma_{s_{1}}$, $h = \eta_{1} \ldots \eta_{s_{2}}$, each $\gamma_{i} \in$ {$g_{1}^{\pm 1}, \ldots , g_{m}^{\pm 1}$} and each $\eta_{i} \in$ {$h_{1}^{\pm 1}, \ldots , h_{r}^{\pm 1}$}. We prove it for commutators of this form by reducing the length of $h$ and then reducing the length of $g$ as follows:

If $s_{2} > 1$,

$[g, \eta_{1} \ldots \eta_{s_{2}}] = [g, \eta_{2} \ldots \eta_{s_{2}}][g, \eta_{1}][g, \eta_{1}, \eta_{2} \ldots \eta_{s_{2}}]$.

If $s_{2} = 1$ and $s_{1} > 1$,

$[\gamma_{1} \ldots \gamma_{s_{1}}, \eta_{1}] = [\gamma_{1}, \eta_{1}][\gamma_{1}, \eta_{1}, \gamma_{2} \ldots \gamma_{s_{1}}][\gamma_{2} \ldots \gamma_{s_{1}}, \eta_{1}]$.
The middle factor on the right hand side can be written as a product of commutators of the desired form by repeatedly applying the equation cited for handling the $s_{2} > 1$ case. Then the final factor has the length of $g$ reduced by 1, as desired. The Commutator Lemma is now proven.

Then $[P,P]$ is generated by a set of such iterated commutators involving $x$, $y$, $x^{-1}$ and $y^{-1}$. Since these have at most $L$ arguments, this gives a generating set of at most $\frac{4^{L+1} - 16}{3}$ elements for $[G,G]$. Then, if one has a $\rho$-element generating set for the subgroup $P_{i}$ in the descending central series of $G$, the Commutator Lemma gives a generating set for $P_{i+1}$ consisting of at most $\frac{(2\rho + 2)^{L+1} - (2\rho + 2)^{2}}{2\rho + 1}$ elements (these finite geometric series arise from summing over the possible lengths of such commutators). Then, continuing in this way, all of the subgroups in the descending central series for $P$ are finitely generated, and the finiteness of $P$ is proven.

Here $\Phi$ denotes the Frattini subgroup.
Denote $|P|$ by $p^{E}$. Let $W_{1}(x,y)$ and $W_{2}(x,y)$ be elements of $P$, considered as words in $x$ and $y$ (well-defined up to the relations). Any substitution $(x,y) \to (W_{1}(x,y),W_{2}(x,y))$ turns relations into other relations (due to the way the relations defining $P$ involve all possible words), so any such substitution defines an endomorphism of $P$. Such a substitution defines an automorphism of $P$ if and only if it is invertible. Since $P$ is finite, an endomorphism from $P$ to itself is invertible if and only if it is surjective. If an endomorphism $\psi$ of $P$ is not surjective, then $\psi(P)$ is contained in some maximal subgroup of $P$ and $\psi(P)\Phi(P)/\Phi(P)$ is a subspace of $P/\Phi(P)$ of positive $\mathbb{Z}/(p)$-codimension. So the endomorphism $\psi$ of $P$ is an automorphism if and only if its action on $P/\Phi(P)$ is via an element of $GL(2,\mathbb{Z}/(p))$. This means that the proportion of endomorphisms of $P$ which are automorphisms is the proportion of 2x2 matrices over $\mathbb{Z}/(p)$ which are invertible. Therefore, we obtain the result that $|Aut(P)| = p^{2E-3}(p^{2}-1)(p-1)$.
Now we let $Aut(P)$ act on the set of non-commuting pairs of elements of $K$. This is well-defined since the relations that hold in $P$ also hold in $K$, and in the subgroup of $K$ generated by two non-commuting elements. To get information about the size of an orbit of a non-commuting pair under this action, it suffices to get information about the order of the stabilizer in $Aut(P)$.
If $a,b \in K$ form a non-commuting pair such that $a = W_{1}(a,b)$ and $b = W_{2}(a,b)$, then $1 = a^{-1}W_{1}(a,b) = b^{-1}W_{2}(a,b)$. $| < a,b > /\Phi(< a,b >)| = p^{2}$, since $< a,b >$ is noncyclic ($< a,b >$ is nonabelian since $(a,b)$ is a non-commuting pair) and generated by 2 elements. This means that, passing to $< a,b >/\Phi(< a,b >)$, we see that $(a,b) \to (W_{1}(a,b),W_{2}(a,b))$ is mapped to the identity matrix in $GL(2,\mathbb{Z}/(p))$. Then likewise, in $Aut(P)$, $(x,y) \to (W_{1}(x,y),W_{2}(x,y))$ must be in the subgroup of order $p^{2E-4}$ which is the kernel of the homomorphism from $Aut(P)$ to $GL(2,\mathbb{Z}/(p))$. Therefore any orbit for the action of $Aut(P)$ on the set of non-commuting pairs of elements of $K$ has size equal to a multiple of $(p^{2}-1)(p^{2}-p)$, and therefore $(p^{2}-1)(p^{2}-p)$ divides $|K|(|K| - c(K))$. Since $|K|$ is a power of $p$, we obtain the congruence $|K| \equiv c(K) \mod{(p^{2}-1)(p-1)}$ as claimed.

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I would just say that to prove $P$ finite, you can simply remark that periodic=locally finite for nilpotent groups. – Steve D Oct 26 '11 at 11:28

Removed- missed "character-free" part. In response to comment below: note that if $p$ is a prime congruent to $\pm 1$ (mod 8), then a finite $p$-group $G$ has $|G| \equiv c(G)$ (mod 32) by an easy adaptation of Burnside's argument.

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