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I am trying to show that the group of Inner Automorphisms are a normal subgroup of the Automorphisms of a group G. So far I have this.

Let $\phi_x:G\rightarrow G$ be the inner automorphism defined as $\phi_x(g)=xgx^{-1} \forall g \in G$. It is clear that $Inn(G)\neq \emptyset$ since $e\in G$ such that $\phi_x(e)=xex^{-1}=e.$ We will show closure and inverses. Observe that $\phi_x\circ \phi_y(g)=\phi_x(ygy^{-1})=xygy^{-1}x^{-1}$.

My question here is for composition of functions, what are you trying to show? Is it just that $\phi_x\circ \phi_y(g)=\phi_{xy}(g)?$.

Now for inverses. (I am trying to follow along with this question here Set of all Inner Automorphisms is a subgroup of the set of all Automorphisms of a group $G$).

So in my case $\phi_e$ is the identity automorphism. Thus $(\phi_x)^{-1}=\phi_{x^{-1}}$. Thus $(\phi_x)^{-1}\phi_h=\phi_{x^{-1}}\phi_h=\phi_{hx^{-1}}=\phi_e\in Inn(G)$. Thus the Inner automorphisms are a subgroup.

Now to show normality. We need to show that $\gamma\circ\phi_x\circ\gamma^{-1}\in Inn(G)$. Since $\gamma\in Aut(G), \gamma$ is a homomorphism. Thus For all $g\in G$

$\gamma\circ\phi_x\circ\gamma^{-1}(g)=(\gamma\circ\phi_x)\gamma^{-1}(g)=\gamma(x\gamma^{-1}(g)x^{-1})=\gamma(x)\gamma(\gamma^{-1}(g))\gamma(x^{-1}=\gamma(x)g\gamma(x^{-1}).$

But this is in the group Inn(G). Thus Inn(G) is a normal subgroup of the Aut(G).

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marked as duplicate by azimut, Rolf Hoyer, Aaron Maroja, egreg, Emily Apr 25 '15 at 23:35

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  • $\begingroup$ See also here. There is also a proof at Wikipedia. $\endgroup$ – Dietrich Burde Dec 7 '14 at 16:56
  • $\begingroup$ For the first part, you want to show that the composition of two inner automorphisms is again an inner automorphism. As you observed, since $y^{-1}x^{-1} = (xy)^{-1}$, you can deduce that $\phi_{x}\circ \phi_{y} = \phi_{xy}$. It looks like you have everything else to me. $\endgroup$ – Alex Wertheim Dec 7 '14 at 16:58