In The number of conjugacy classes and the order of the group, a key step in the proof of the congruence saying that $G$ being a $p$-group means that $|G| \equiv c(G) \mod{(p^{2}-1)(p-1)}$ is the construction of a finite group (denoted $P$ in that post) whose relations, as given, are satisfied by \emph{every} word in the generators.

Let $Q$ denote a finite $p$-group which has a presentation of the kind possessed by $P$. More precisely, suppose $Q$ is generated by $x_{1}, \ldots, x_{r}$ and that the relations defining $Q$ are given in infinite families, where each family is of the form $w( e_{1}, \ldots, e_{L} ) = 1$, where $w$ is some word in the free group on $L$ generators and $e_{1}, \ldots, e_{L}$ vary independently over all words in $x_{1}, \ldots, x_{r}$.

The reasoning from before should apply (though I cannot fill in all gaps) to describe $\mathrm{Aut}(Q)$ (and obtain its order):
Since $Q$ is finite, an endomorphism $\phi: Q \to Q$ is injective iff it is surjective. So an endomorphism of $Q$ is an Automorphism iff it is surjective. If $\phi(x_{1}), \ldots, \phi(x_{r})$ do not generate all of $Q$, then the subgroup they generate is contained (since $Q$ is finite) in some maximal subgroup $M$ of $Q$. $M$ contains the Frattini subgroup $\Phi(Q)$, so this endomorphism fails to be surjective when it is converted to an endomorphism of the maximal elementary abelian quotient $Q/ \Phi(Q)$.
Gap 1: I do not immediately see how to prove that $|Q/ \Phi(Q)| = p^{r}$. It is clear that, since $Q$ is generated by an $r$-element set, $|Q/ \Phi(Q)| \leq p^{r}$. The nature of the relation set should imply that $Q$ is trivial if $|Q/ \Phi(Q)| < p^{r}$.
Assuming Gap 1 is fillable, when the words expressing $\phi(x_{1}), \ldots, \phi(x_{r})$ are written in terms of $x_{1}, \ldots, x_{r}$, one can form the matrix whose $(i,j)$ entry is the sum of the exponents, regarded as an element of $\mathbb{Z}/(p)$, to which $x_{i}$ appears in $\phi(x_{j})$.
Gap 2: I do not immediately see how to prove that element of $\mathbb{Z}/(p)$ is well-defined.
Assuming Gap 2 is also fillable, the matrix thus obtained will be invertible iff $\phi$ is surjective, and thus an \mathrm{Aut}omorphism of $Q$. Since all $r \times r$ matrices over $\mathbb{Z}/(p)$ arise this way from endomorphisms of $Q$ (assuming Gap 1 is filled), construction of this matrix yields a surjective homomorphism from $\mathrm{Aut}(Q)$ to $\mathrm{GL}(r,\mathbb{Z}/(p))$. What is the kernel of this homomorphism?
The kernel consists of all preimages of the identity matrix under that abelianization modulo $p$ map. If $|Q| = p^{E}$, assuming Gap 1 is filled, $|\Phi(Q)| = p^{E-r}$. Then any kernel element can be obtained by sending $x_{i}$ to $x_{i}f_{i}$ for all $i$, where $f_{i} \in \Phi(Q)$. This means that the order of the kernel is $p^{r(E-r)}$, and that $|\mathrm{Aut}(Q)| = p^{r(E-r)} \Pi_{k=0}^{r-1} (p^{r}-p^{k})$.

Going back to this question for the quaternion group Q_{8}, $\mathrm{Aut}(Q_{8}) \cong S_{4}$ so $\mathrm{Aut}(Q_{8})$ has $\mathrm{GL}(2, \mathbb{Z}/(2) ) \cong S_{3}$ as a quotient. The kernel of this quotient map consists of the automorphisms of $Q_{8}$ which send each element of $Q_{8} \setminus \Phi(Q_{8})$ to itself, except possibly for sign. Since { $\pm{1}$ } $ = \Phi(Q_{8})$, this sounds exactly like what was covered in the discussion of $\mathrm{Aut}(Q)$ above.
Yet I still have trouble obtaining a presentation of the kind $Q$ satisfies for the quaternion group of order 8:

One may say that, in $Q_{8}$, all elements have order dividing 4: $e^{4} = 1$ for all words $e \in \langle x,y \rangle$.
One may also say that, in $Q_{8}$, all elements square to an element of the center: $e_{1}^{2}e_{2} = e_{2}e_{1}^{2}$ for all words $e_{1}, e_{2} \in \langle x,y \rangle$.

These two statements do not capture $Q_{8}$ completely, since the group with those relations turns out to have order $32$:

$T = \langle x,y| e^{4}=1, e_{1}^{2}e_{2} = e_{2}e_{1}^{2} \rangle$ is nonabelian, since the nonabelian group $Q_{8}$ is one of its quotients.
Now consider the element $x^{2}y^{2}(xy)^{2} \in T$. This element is clearly a product of squares, so it is a product of central elements of order dividing 2. It therefore has order dividing 2 itself.
Yet $T/\langle x^{2}y^{2}(xy)^{2} \rangle$ is abelian, since $x^{2}y^{2}(xy)^{2} = 1$ implies $x^{2}y^{2} = (xy)^{-2})$, which implies (since every element has order dividing 4) $x^{2}y^{2} = (xy)^{2} $, which implies (left-multiplying by $x^{-1}$ and right-multiplying by $y^{-1}$) $xy = yx$.
Therefore $x^{2}y^{2}(xy)^{2}$ is a nontrivial element of $T$, so it has order 2. $T/\langle x^{2}y^{2}(xy)^{2} \rangle$ is, therefore, the largest 2-generated abelian group of exponent 4 (and being abelian subsumes the centrality of squares). So $T/\langle x^{2}y^{2}(xy)^{2} \rangle \cong C_{4} \times C_{4}$ and it is now clear that $|T| = 32$.

The question indicated by the title asks whether more relations can be added, in this symmetric fashion, to $T$ to obtain $Q_{8}$. The natural extension of this question is whether or not there exists a 'nice' criterion for looking at an arbitrary $p$-group and determining whether or not it has a presentation of the kind discussed here. Clearly the discussion of $\mathrm{Aut}(Q)$ gives some preliminary necessary conditions for the existence of such a presentation.

  • $\begingroup$ All of your discussion about automorphisms of p-groups was done by Philip Hall, and is given in Marshall Hall's group theory text, for example. In particular, there are no gaps. You can prove this yourself if you recall that $\Phi(G)$ is the set of "non-generators", and that in p-groups, $\Phi(G)=G^p[G,G]$. $\endgroup$
    – Steve D
    Jan 30, 2012 at 16:47
  • 1
    $\begingroup$ ALso, it would seem the quaternion group does not have such a presentation. Because clearly, if it did, we could add the relators which make up $T$ without changing the group, so we can just assume that we start with $T$ and want to add more symmetric relators. Now there is only one subgroup $N$ of $T$ such that $T/N$ is $Q_8$. If that subgroup of $T$ was generated by symmetric relators, $N$ would need (at the least) to be fully invariant; however it is not. I checked all this in GAP. $\endgroup$
    – Steve D
    Jan 30, 2012 at 17:02

1 Answer 1


A group given by your presentation for words $w_1,...,w_n$ is called relatively free in the variety of groups given by the identities (laws) $w_1=1,...,w_n=1$. For a characterization of relatively free groups see Hanna Neumann's book "Varieties of groups" . The characterization says that $G=\langle X\rangle$ is relatively free with free generating set $X$ iff every map $\phi\colon X\to H$ into a group $H$ satisfying the same identities as $G$ can be extended to a homomorphism $G\to H$ (the extension is then unique). The variety generated by the quaternion group $Q_8$ (i.e. the smallest variety containing that group) is also described in Hanna Neumann's book. It is defined by three identities $[x^2,y]=1, [[x,y],z]=1, x^4=1$ (the second identity actually follows from the other two and is redundant). Also $Q_8$ is 2-generated, so if it was relatively free, its rank would be 2. But the dihedral group of order 8 generates the same variety (see the book), and all homomorphisms from $Q_8$ into $D_8$ have Abelian images. Thus $Q_8$ is not relatively free.


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