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Find Aut$(G)$, Inn$(G)$ and $\dfrac{\text{Aut}(G)}{\text{Inn}(G)}$ for $G = \mathbb{Z}_2 \times \mathbb{Z}_2$.

Here is what I have here:

Aut$(G)$ consists of 6 bijective functions, which maps $G$ to itself, since Aut$(\mathbb{Z}_2 \times \mathbb{Z}_2) \approx S_3$.

I think the next part goes wrong. For Inn$(G)$, letting $\kappa_x : G \rightarrow G$ to be the conjugate function, I found $\{e, \kappa_{(0,1)}, \kappa_{(1,0)}, \kappa_{(1,1)}\}$.

I can't determine $\dfrac{\text{Aut}(G)}{\text{Inn}(G)}$ yet since I need to correctly determine Inn$(G)$.

Any advices or comments?

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    $\begingroup$ Something definitely goes wrong, because the order of a subgroup must divide the order of the group. $\endgroup$ May 18 '13 at 15:33
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    $\begingroup$ What do inner automorphism groups do on abelian groups? Just think about how they are defined. $\endgroup$ May 18 '13 at 15:36
  • $\begingroup$ Use N/C lemma. $\endgroup$
    – Mikasa
    May 18 '13 at 15:42
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Hint: to expand on Tyler's comment, can you prove that, more generally, $\operatorname{Inn}(G) = \{0\}$ iff $G$ is abelian?

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  • $\begingroup$ Oh... So since $G$ is indeed abelian, Inn$(G)$ is trivial. What about nonabelian group? I believe Inn$(G)$ is not trivial, am I right? $\endgroup$
    – NasuSama
    May 18 '13 at 16:16
  • $\begingroup$ @NasuSama: Did you show that $Aut(Z_2\times Z_2)\cong S_3$? $\endgroup$
    – Mikasa
    May 18 '13 at 16:26
  • $\begingroup$ Actually, I did. $\endgroup$
    – NasuSama
    May 18 '13 at 16:38
  • $\begingroup$ @NasuSama in my answer there is the general case about Inn($G$) in a group not necessarily abelian :) $\endgroup$
    – Riccardo
    May 18 '13 at 16:41
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First note that $G$ is abelian, therefore $Z(G) = G$
($Z$ is the Center of $G$).

Then you can use a little proposition (Humphreys pg.73-74) that tells you $\text{Inn}(G) \approx \frac{G}{\text{Z}(G)}$

The proof of this prop in a few words:

define an automorphism (prove it) $\varphi_x(g)=xgx^{-1}$, then define $\phi : G \to \text{Aut}(G)$ by $x \mapsto \varphi_x$ and find its image and its kernel, then apply first theorem of homomorphism and you conclude.

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  • $\begingroup$ In this particular Abelian case, I don't think it takes any proposition to realize that $\phi_g(x)=gxg^{-1}=gg^{-1}x=x$ for every inner automorphism $\phi_g$. Even though the general statement you mention is good to know. $\endgroup$
    – Julien
    May 18 '13 at 15:51
  • $\begingroup$ yes yes, the reason is I was reading the chapters after this in the Humphreys, and I have this prop fresh in mind :) $\endgroup$
    – Riccardo
    May 18 '13 at 15:54

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