TITLE: Classifying the groups of order 56
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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.