What does it mean for an element of a group to be a function on the group?

Question

I am trying to learn group theory for my physics research. I'm working through these notes: http://www.damtp.cam.ac.uk/user/ho/GNotes.pdf

On page 3, the author states:

"It is straightforward to see that the set of all automorphisms of $G$ itself forms a group $\operatorname{Aut} G$ which must include $G/\mathcal{Z}(G)$ as a normal subgroup."

This confuses me; there seems to be a type error. An element of $\operatorname{Aut} G$ is a function $\phi: G \rightarrow G$. An element of $G/\mathcal{Z}(G)$ is an element of $G.$ So the latter group cannot be a subset of the former group, hence the author must be assuming some identification between group elements and functions on group elements.

What identification is the author likely assuming? He references defining what he calls an "inner automorphism" by "$\phi_g (g_i) = g g_i g^{-1},$" and this seems like the identification I'm looking for, but I am uncomfortable with the fact that he's called this by the name "inner automorphism."

Answer 1

Note: Each element $a \in G$ defines an automorphishm $$\phi_a : G \to G \,;\, \phi_a(x)=axa^{-1}$$

The mapping $$F : G \to Aut(G) \,;\, F(a)=\phi_a$$ is a group homomorphism. Therefore, by the first isomorphism Theorem, we get an isomorphism: $$F : G / \ker(F) \to Im(F) \leq Aut(G)$$ and it is straight forward to check $\ker(F)=\mathcal{Z}(G)$.

Now, via this "natural" mapping, $G/\mathcal{Z}(G)$ can be identified (is isomorphic) with a subgroup of $Aut(G)$, and we usually consider them to be "the same".

Answer 2

It would be better

which includes a normal subgroup isomorphic to $G/\mathcal{Z}(G)$

Given $g\in G$, you can consider the inner automorphism $\phi_g$ defined by $$ \phi_g(x)=gxg^{-1} $$ It is easy to see this is indeed an automorphism of $G$. Moreover $$ \phi_g\circ\phi_h=\phi_{gh} $$ so the map $\phi\colon G\to\operatorname{Aut}G$ defined by $g\mapsto \phi_g$ is a group homomorphism. Its kernel is $\mathcal{Z}(G)$, so the homomorphism theorem says we have an injective homomorphism $$ G/\mathcal{Z}(G)\to\operatorname{Aut}G $$ It remains to show the image is a normal subgroup.

Answer 3

You're right: it results from this identification. More precisely, any group acts on itself by inner automorphisms (one also says by conjugation).

There is a canonical homomorphism $\begin{aligned}[t]\varphi \colon G&\longrightarrow \operatorname{Aut}G\\g&\longmapsto(x\mapsto gxg^{-1})\end{aligned}$ The kernel of this homomorphism is, by definition, the centre of $G$, $Z(G)$, so that we obtain an embedding $\;G/Z(G)\hookrightarrow \operatorname{Aut}G$.

The image (the inner automorphisms) are a normal subgroup of $\operatorname{Aut}G$, because one easily checks that, if $f$ is an automorphism $G$, the conjugate of the inner automorphism induced by $g$ is the inner automorphism induced by $\varphi(g)$.