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How to find $\left|\operatorname{Aut}(\mathbb Z_2\times\mathbb Z_2)\right|=?$

Since $\mathbb Z_2\times\mathbb Z_2$ has 4 elements, $1 \leq \left|\operatorname{Aut}(\mathbb Z_2\times\mathbb Z_2)\right|\leq \left|S_4\right|$.

How to proceed further?

Thanks in advance...

Thomas Andrews
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Mirunalini_UML
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1 Answers1

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Notice that in $(\mathbb Z /2\mathbb Z) \times (\mathbb Z /2\mathbb Z)$ all elements besides the neutral have order two (and each of them is just the product of the other two!), so every automorphisms "permutes" them and maps the neutral element to itself.

We have therefore shown something stronger: $\textrm{Aut} (\mathbb Z /2\mathbb Z) \cong S_3$.

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    You have not quite argued that all the permutations are actually automorphisms. – Tobias Kildetoft Apr 20 '15 at 12:30
  • We know that $e$ must be fixed, so there are at most $6$ automorphisms. Can't I define $S_3$ as group of permutations of ${(0,1),(1,1),(1,0)}$? –  Apr 20 '15 at 12:32
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    Yes, but it is a bit of a coincidence that all of those permutations are actually automorphisms. – Tobias Kildetoft Apr 20 '15 at 12:34
  • For more on this phenomenon, see http://math.stackexchange.com/questions/452596/is-a-bijection-from-a-group-to-itself-automatically-an-isomorphism-if-it-maps-th/ – Tobias Kildetoft Apr 20 '15 at 12:35