# Order of the automorphism group of $(\mathbb{Z}/2\mathbb{Z})^3$

I'm trying to find the number of distinct automorphisms of the group $\mathbb{Z}/2\mathbb{Z} × \mathbb{Z}/2\mathbb{Z} × \mathbb{Z}/2\mathbb{Z}$. In the case of $Aut(\mathbb{Z}/2\mathbb{Z} × \mathbb{Z}/2\mathbb{Z})$, it's easy to see that every permutation of elements preserves the structure of the group operation; however, this is clearly not the case for the more complex group $Aut (\mathbb{Z}/2\mathbb{Z} × \mathbb{Z}/2\mathbb{Z} × \mathbb{Z}/2\mathbb{Z})$ (Consider, for example, the permutation that maps $(0,1,1) \mapsto (0,1,0)$ and $(1,0,1) \mapsto (1,1,0)$, where $(0,1,0) \neq (1,0,0)$).

I'm wondering how to neatly determine which permutations of $\mathbb{Z}/2\mathbb{Z} × \mathbb{Z}/2\mathbb{Z} × \mathbb{Z}/2\mathbb{Z}$ are homomorphic.

• $(1,0,0)$ can be mapped to any of the $7$ nonzero elements. After choosing that image there are only $6$ possible targets for $(0,1,0)$. If these first two images are $g$ and $h$, then $(0,0,1)$ cannot be mapped to $0,g,h$ or $g+h$,so it has $4$ possible images. – Derek Holt Oct 13 '16 at 21:45
• It's much easier to first think about the endomorphisms of this group, which form a ring, namely the ring $M_3(\mathbb{Z}/2\mathbb{Z})$ of $3 \times 3$ matrices over $\mathbb{Z}/2\mathbb{Z}$. The automorphism group is then the group of units of this ring, which is the group $GL_3(\mathbb{Z}/2\mathbb{Z})$ of invertible $3 \times 3$ matrices over $\mathbb{Z}/2\mathbb{Z}$. You can count the number of elements of this group by counting how many possibilities there are for each column, which is another way of describing Derek's argument. – Qiaochu Yuan Oct 13 '16 at 22:08
• This is implicit in Qiaochu's comment, but $(\mathbb{Z}/2\mathbb{Z})^3$ is a vector space over $\mathbb{F}_2 = \mathbb{Z}/2\mathbb{Z}$ and you can show every abelian group homomorphism is a $\mathbb{F}_2$-linear map, and conversely. Then you can use your knowledge of linear algebra to solve the problem. – André 3000 Oct 14 '16 at 0:30