# If the group if abelian, must all automorphisms be outer automorphisms?

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.

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

For abelian group $G$, we know $\,\operatorname{Inn}(G) \cong e.\;$ So, we have that $$\operatorname{Out}(G)\cong \operatorname{Aut}(G)/\operatorname{Inn}(G)\cong \operatorname{Aut}(G)$$
• Yup! Another question: What if $G$ is nonabelian, say $D_4$? How would $Out(D_4)$ look like? – NasuSama May 21 '13 at 17:54
• @NasuSama: this post might clear up confusion about $D_4$ – Namaste May 21 '13 at 18:30