IMC 2008 · Day 1 · P5

Yes. Let $H$ be the commutative group $H = \mathbb{F}_2^3$, where $\mathbb{F}_2 \cong \mathbb{Z}/2\mathbb{Z}$ is the field with two elements. The group of automorphisms of $H$ is the general linear group $GL_3 \mathbb{F}_2$; it has $$(8 - 1) \cdot (8 - 2) \cdot (8 - 4) = 7 \cdot 6 \cdot 4 = 168$$ elements. One of them is the shift operator $\varphi : (x_1, x_2, x_3) \mapsto (x_2, x_3, x_1)$.

Now let $T = \{a^0, a^1, a^2\}$ be a group of order 3 (written multiplicatively); it acts on $H$ by $\tau(a) = \varphi$. Let $G$ be the semidirect product $G = H \rtimes_\tau T$. In other words, $G$ is the group of 24 elements $$G = \{ b a^i : b \in H,\ i \in (\mathbb{Z}/3\mathbb{Z}) \}, \qquad ab = \varphi(b) a.$$ $G$ has one element $e$ of order 1 and seven elements $b$, $b \in H$, $b \ne e$ of order 2.

If $g = ba$, we find that $g^2 = baba = b \varphi(b) a^2 \ne e$, and that $$g^3 = b \varphi(b) a^2 b a = b \varphi(b) a \varphi(b) a^2 = b \varphi(b) \varphi^2(b) a^3 = \psi(b),$$ where the homomorphism $\psi : H \to H$ is defined as $\psi : (x_1, x_2, x_3) \mapsto (x_1 + x_2 + x_3)(1, 1, 1)$. It is clear that $g^3 = \psi(b) = e$ for 4 elements $b \in H$, while $g^6 = \psi^2(b) = e$ for all $b \in H$.

We see that $G$ has 8 elements of order 3, namely $ba$ and $ba^2$ with $b \in \operatorname{Ker} \psi$, and 8 elements of order 6, namely $ba$ and $ba^2$ with $b \notin \operatorname{Ker} \psi$. That accounts for orders of all elements of $G$.

Let $b_0 \in H \setminus \operatorname{Ker} \psi$ be arbitrary; it is easy to see that $G$ is generated by $b_0$ and $a$. As every automorphism of $G$ is fully determined by its action on $b_0$ and $a$, it follows that $G$ has no more than $$7 \cdot 8 = 56$$ automorphisms.

Remark. $G$ and $H$ can be equivalently presented as subgroups of $S_6$, namely as $H = \langle (12), (34), (56) \rangle$ and $G = \langle (135)(246), (12) \rangle$.