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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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- Need to prove that (S,*) defined by the binary operation a*b = a+b+ab is an abelian group on S = R \ {1}

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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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