I am having some thought about this question.

Clearly, if the group $G$ is abelian, then $\mathrm{Inn}(G) = {e}$. But what about $\mathrm{Aut}(G)$ and $\mathrm{Out}(G) = \mathrm{Aut}(G)/\mathrm{Inn}(G)$? Must $\mathrm{Aut}(G) = \mathrm{Out}(G)$? Take $G = \mathbb{Z}_2 \times \mathbb{Z}_2$ as the example. We know that $\mathrm{Aut}(G)$ consists of 6 bijective functions. Must $\mathrm{Out}(G)$ also consist of 6 bijective functions?

Different example:

$G = D_4$

$G$ is not abelian and noncyclic. It consists of $4$ elements in $\mathrm{Inn}(G)$ and $8$ elements in $\mathrm{Aut}(G)$. Must $\mathrm{Out}(G)$ consist of 2 elements? I found that there are 4 functions instead of 2.

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    $\begingroup$ If $G$ is any group then $G$ is isomorphic to $G/\{1\}$. $\endgroup$ – Tobias Kildetoft May 21 '13 at 17:47
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    $\begingroup$ You have answered your own question. Since $\operatorname{Inn}(G)\cong e$, it follows that $\operatorname{Out}(G)\cong \operatorname{Aut}(G)/\operatorname{Inn}(G)\cong \operatorname{Aut}(G)$. $\endgroup$ – Jared May 21 '13 at 17:48
  • $\begingroup$ @Jared What about nonabelian group like $D_4$? $\endgroup$ – NasuSama May 21 '13 at 17:52

Can you see that you have correctly answered your question?

For abelian group $G$, we know $\,\operatorname{Inn}(G) \cong e.\;$ So, we have that $$\operatorname{Out}(G)\cong \operatorname{Aut}(G)/\operatorname{Inn}(G)\cong \operatorname{Aut}(G)$$

  • $\begingroup$ Yup! Another question: What if $G$ is nonabelian, say $D_4$? How would $Out(D_4)$ look like? $\endgroup$ – NasuSama May 21 '13 at 17:54
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    $\begingroup$ @NasuSama, please ask a new question. Asking about non-abelian groups is a question whose title clearly states that it is about abelian groups is not a great idea! $\endgroup$ – Mariano Suárez-Álvarez May 21 '13 at 17:56
  • $\begingroup$ @NasuSama: this post might clear up confusion about $D_4$ $\endgroup$ – Namaste May 21 '13 at 18:30
  • $\begingroup$ Nice work Amy + $\endgroup$ – mrs May 21 '13 at 18:39
  • $\begingroup$ NasuSama: did this help at all? (Nice answer, btw, (+1) on the rational root theorem!) $\endgroup$ – Namaste May 22 '13 at 0:39

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