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So I am given an abelian group of order $8$ such that for all non-identity elements $x^2 = e$ (all elements have order two). So I know the answer is gonna be $168$, but I gotta prove this.

So far I have that there are exactly $7$ subgroups of order $4$, and each one has $6$ automorphisms. That gives me a total of $42$ automorphisms. Now there must be a justification for multiplying this number by $4$ to get the $168$. But I can't seem to find a justification for this.

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Note that since all elements have order $2$, the group is a vector space over the field with $2$ elements, so the automorphism group consists of invertible linear transformations, ie, invertible matrices over the field with $2$ elements. – Tobias Kildetoft Mar 13 '13 at 1:20
Dear Geano: Welcome! Thank you for posting a good question: one which you clearly specify, and in which you address what you've figured out so far, where you're stuck, etc. For any post, that's refreshing for us to see. And for being your first post here, that is really nice to see. +1 – amWhy Mar 13 '13 at 1:37
Yeah! Hope you stay @Geono. – Alexander Gruber Mar 13 '13 at 1:46
i just discovered this web page, and i am definitely staying:) – Geano Mar 13 '13 at 1:56

If you are familiar with finite fields, it's easiest to see this by viewing $G=\mathbb{Z}_2\times \mathbb{Z}_2 \times \mathbb{Z}_2$ as a vector space of dimension $3$ over $\mathbb{F}_2$. In this way, we see that the automorphism group of $G$ is $GL_3(\mathbb{F}_2)$, the group of $3 \times 3$ invertible matrices over $\mathbb{F}_2$. By counting the ways to make three linearly independent vectors, we see that $$|\operatorname{Aut}(G)|=(2^3-1)(2^3-2)(2^3-2^2)=168.$$ If you don't know what a finite field is, this isn't as easy. Finite fields are easy to understand if they have prime order, though - just add and multiply the numbers $0,\ldots, p-1$ modulo $p$. (As it turns out, finite fields always have prime power order, but things get more complicated when the power is greater than $1$.)

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it seems this method is the most noted, but we haven't gotten into finite fields yet. – Geano Mar 13 '13 at 1:27
@Geano This really is the best way to understand it. The finite field $\mathbb{F}_2$ is really easy: it only has two elements, $0$ and $1$, with the operations $0+1=1$, $1+1=0$, $1\cdot 1=1$, $0\cdot 1 = 0$. – Alexander Gruber Mar 13 '13 at 1:29
my thoughts on the matter was that each of the 6 automorphisms, on a specific 4 sub group leaves 4 elements untouched...and this is where we are stuck – Geano Mar 13 '13 at 1:30
@Geano The multiplication by $4$ comes exactly from the possibilities for the third column in the linear transformations. Perhaps you can think of the subgroups of order $4$ as permutable subspaces. – Alexander Gruber Mar 13 '13 at 1:32
Alexander, maybe you could guide me in the right direction in this line of thought. So V = Z2 X Z2 X Z2 => It forms a 3D cube. Now (help me here) an automorphism has to preserve the spatial position of the cube. So there is seven ways to pick basis vector 1, 6 ways to pick basis vector...and all automorphisms would be described this way? – Geano Mar 13 '13 at 5:09

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