Take the 2-minute tour ×  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.  share|improve this question 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