I am trying to classify the groups of order $56=2^3\cdot 7$. I think I have successfully eliminated most of the cases, and I am left with the last: When the group of order 8 is normal and it is isomorphic to $Z_2\times Z_2\times Z_2$. So, we seek homomorphisms $\varphi:Z_7\to \mathrm{Aut}(Z_2\times Z_2\times Z_2)\cong \mathrm{GL}(3,\mathbb{Z}_2)$.

$\varphi$ is completely determined by its action on $1\in Z_7$, and since its order is 7, a prime, its image, $\varphi(1)$, must have order 1 or 7. The former corresponds to the trivial product $Z_2\times Z_2\times Z_2\times Z_7$. The interesting one would arise from the latter, and here is where I am lost. Yes, $|\mathrm{GL}(3,\mathbb{Z}_2)|=168=2^3\cdot 3\cdot 7$, so by Cauchy, the existence of such an element (of order 7) is guaranteed. But how do I find it?

In general, I seek advice on problems like this where the automorphism group is given by the general linear group. From here and there, I saw people using eigenvalues, but I am not familiar with that technique, given my training. If someone could be so kind as to explain that method in detail, it would be very greatly appreciated.

  • 2
    $\begingroup$ Probably the easiest way to produce an element of order $7$ in ${\rm GL}(3,2)$ is to let $\omega$ be a generator of the multiplicative group of the finite field ${\mathbb F}_8$ of order $8$, and regard ${\mathbb F}_8$ as a vector space of dimension $3$ over ${\mathbb F}_2$. Then multiplication by $\omega$ induces an element of order $7$ in ${\rm GL}(3,2)$. A more boring method is to just choose random $3 \times 3$ matrices over ${\mathbb F}_2$. There are $48$ element sof order $7$, so you will find one quite quickly. $\endgroup$
    – Derek Holt
    Jul 13, 2018 at 20:41
  • $\begingroup$ By the way, you will want to apply the Sylow conjugacy result (to $GL(3,2)$) to show that there is (up to equivalence) only one way for an order $7$ group to act by automorphisms on $2^3$. $\endgroup$
    – C Monsour
    Jul 16, 2018 at 0:14

2 Answers 2


A preliminary remark (which I think you understand): there is exactly one way to make the group $(\mathbb{Z}_p)^n=\mathbb{Z}_p \oplus\cdots\oplus \mathbb{Z}_p $ (I write it additively) into a vector space over the field of $p$ elements $\mathbb{F}_p$: for the axioms to be true you must set $k\cdot x=x+x+\cdots+x$, the sum of $k$ $x$'s. This means that the group automorphisms are also vector space automorphisms, and so the automorphism group is exactly $\text{GL}(n,\mathbb{F}_p)$.

I think that if you want to tackle questions like this you do need to know something about Linear Algebra: the theorem of central importance here is (I suggest) the Rational Canonical Form Theorem. I think it is a good idea to understand its use first, before worrying about eigenvalues: that often involves extending the field (as in @Derek Holt 's comment) and that is not quite so intuitive.

In the case you are considering you want to find an $\alpha:V_3(\mathbb{F}_2)\to V_3(\mathbb{F}_2)$ which is of order $7$.

You know, then, that $\alpha^7=1$, and so the minimal polynomial $m_{\alpha}(X)$ of $\alpha$ must divide $X^7-1$.

Over $\mathbb{F}_2$ we can calculate easily that $X^7-1=(X-1)(X^3+X^2+1)(X^3+X+1)$, and both the cubic factors must be irreducible.

Disregarding the trivial case we must have that $m_{\alpha}(X)=X^3+X^2+1$ or $X^3+X+1$. The Theorem of the Rational Canonical Form will now tell us that every $\alpha$ is conjugate to the companion matrix of one or other of these cubics. That is we may, by choosing a suitable basis of $V_3(\mathbb{F}_2)$, assume that $\alpha$ is one of:

$$ \begin{pmatrix} 0 & 0 & 1\\ 1& 0& 0\\ 0 & 1 & 1\\ \end{pmatrix} \text{ or } \begin{pmatrix} 0 & 0 & 1\\ 1& 0& 1\\ 0 & 1 & 0\\ \end{pmatrix}. $$

Note that if $\alpha$ satisfies $X^3+X^2+1$ then $\alpha^{-1}$ satisfies the other irreducible, $X^3+X+1$. So in considering non-trivial group extensions of the elementary abelian group of order $8$ by a cyclic group of order $7$ there is really only one possibility: $$ \langle x,y,z,s |x^2=y^2=z^2=[x,y]=[x,z]=[y,z]=1, s^7=1, x^s=y, y^s=z, z^s=xz \rangle. $$

(Exercises for the reader: find all groups of order $80$ with an elementary abelian normal subgroup of order $16$, and all groups of order 351 with an elementary abelian subgroup of order $27$.)


Viewing the 3-dimensional vector space over $\mathbb{F}_2$ as the field extension $\mathbb{F}_2\leq \mathbb{F}_8$ is very advantegous in this problem. You are looking for a matrix that is a root of the polynomial $x^7-1$, or to make the connection more apparent, a root of the polynomial $x^8-x$. Roots of this polynomial are exactly the elements of $\mathbb{F}_8$. So Factorize the polynomial into irreducible components:

$x^8-x= x(x-1)(x^3+x+1)(x^3+x^2+1)$. This is easy to find: we know that we are looking for two irreducible polynomials of degree 1 and two of degree 3 by general theory of finite fields.

So an easy way to construct a $3\times 3$ matrix over $\mathbb{F}_2$ of order 7 is to produce the companion matrix of the polynomial $x^3+x+1$, for example.

So one solution is: $\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$

Clearly, there is exactly one non-Abelian group of order 56 with an elementary Abelian normal subgroup of order 8. As you pointed out, we are looking for a nontrivial homomorphism $\varphi: \mathbb{Z}_7\rightarrow GL(3,2)$. You can construct such a homomorphism using the above matrix, for example, so existence is clear. As for uniqueness, you need two observations.

  • If two such homomorphisms have the same image, say $Im(\varphi)=Im(\psi)$, then one can be obtained from the other by composition with an automorphims from the right: $\varphi=\psi\circ \alpha$, $\alpha\in Aut(\mathbb{Z}_7)$.

  • All 7-element subgroups of $GL(3,2)$ are conjugate. So given $\varphi, \psi$, you can make their images coincide by composition with an automorphism from the left: $Im(\varphi)= Im(\beta\circ \psi)$, $\beta\in Aut(GL(3,2))$.

Composition with automorphisms yield isomorphic semidirect products.


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