After a week in my Abstract Algebra class, the professor proposed this as a problem. I'm not entirely sure where to begin. $ V = \{ e, \tau, \tau_1, \tau_2 \}$, so I'm not sure exactly what is meant by $\text{Aut} (V)$. Is it simply saying that the only way to have an automorphism on $V$ is to rearrange the order of the elements of $V$ and therefore $\text{Aut} (V) \cong S_3$?
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Any automorphism $\phi$ of $V$ sends its identity element $e$ to itself. What is interesting is the way that $\phi$ rearranges the other three elements. The claim here is that any of the six possible bijections from $\tau, \tau_1, \tau_2$ to itself is in fact a group automorphism of $V$. This will give an isomorphism from $\operatorname{Aut}(V) \cong S_3$ that sends an automorphism $\phi \in \operatorname{Aut}(V)$ to its corresponding permutation of the three elements $\tau, \tau_1, \tau_2$.
Elchanan Solomon
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1Why is it $\operatorname{Aut}(V) \cong S_4$ not $\operatorname{Aut}(V) \cong S_3$? – Anthony Peter Oct 27 '13 at 03:18
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Because I'm an idiot who makes mistakes. Sorry. – Elchanan Solomon Oct 27 '13 at 03:22
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Could you formally write that $\forall \Gamma (\phi), \exists ! \sigma(S_3) : \sigma(S_3) \cong \Gamma(\phi)$?, where $\Gamma$ is the isomorphism and $\sigma$ is an arbitrary permutation – Anthony Peter Oct 27 '13 at 18:29
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There are three elements of order $2$ in $V$. Identity element must go to identity element by any automorphism, and you can permute in any way the elements of order $2$, so $Aut(V)=S_3$.
Of course, you need to check that any permutation gives you an automorphism. It is obviously bijective, and is very easy to check directly that any such permutation preserves the group law.
Sasha Patotski
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