I am reading Abstract Algebra: 3rd Edition by Dummit and Foote. On page 134, example 1, it says:
A group $G$ is abelian if and only if every inner automorphism is trivial. If $H$ is an abelian normal subgroup of $G$ and $H$ is not contained in $Z(G)$, then there is some $g\in G$ such that conjugation by $g$ restricted to $H$ is not an inner automorphism of $H$.
I'm trying to prove both of these statements, and I'm having trouble understanding the second one.
The first statement is pretty easy: $$\begin{align*} G\text{ is abelian}&\iff gh=hg\text{ for all }g,h\in G\\ &\iff ghg^{-1}=h\text{ for all }g,h\in G\\ &\iff\sigma_g(h)=h\text{ for all }\sigma_g\in \text{Inn}(G), h\in H\\ &\iff \sigma_g = \text{id}_{\text{Aut}(G)}\text{ for all }\sigma_g\in \text{Inn}(G). \end{align*}$$
(Is this correct?)
Next, I suppose I don't understand the second statement. What I understand it to mean is that if $H$ is an abelian normal subgroup of $G$ and $H\not\subseteq Z(G)$, then there exists $g\in G$ such that $\sigma_g:h\mapsto ghg^{-1}$ is not an automorphism of $H$.
What I've tried:
Since $H\unlhd G$, we have that $h_1g=gh_2$ for some $h_1, h_2\in H$, for all $g\in G$.
Then $h_1 = gh_2g^{-1}=\sigma_g(h_2)$.
This doesn't seem like it gets me anywhere, since it just shows that for some of the elements $h$ of $H$, we have that $\sigma_g(h)\in H$.
This leads me to take an arbitrary $\sigma_g(h)$ ($g\in G, h\in H$ are arbitrary).
But then $\sigma_g(h) = ghg^{-1} = gg^{-1}h' = h'\in H$ for some $h'\in H$ (by normality), so clearly $\sigma_g$ always maps elements in $H$ to elements in $H$.
Also, $\sigma_g(h_1h_2) = gh_1h_2g^{-1} = gh_1g^{-1}gh_2g^{-1} = \sigma_g(h_1)\sigma_g(h_2)$ so $\sigma_g$ is always a homomorphism from $H$ to $H$ (when restricted to $H$).
What am I missing? Am I misunderstanding the statement? When is $\sigma_g$ ever not an inner automorphism? How do I use the fact that $H\not\subseteq Z(G)$?
The first part has a couple of slight errors: your first few implications have $h,g\in G$, but your last two have $h\in H$. Remember: you are trying to prove that $G$ is abelian if and only if every inner automorphism of $G$ is trivial. There is no $H$ yet. So you may want to change those $g$ and $h$ to $x$ and $y$ to avoid confusion. And it should be $\mathrm{id}_G$, not $\mathrm{id}_{\mathrm{Aut}(G)}$: the function is the identity automorphism of $G$, not of the automorphism group. So it should read something like the following (note the boldface, which is where your errors are):
$$\begin{align*} G\text{ is abelian}&\iff xy=yx\text{ for all }x,y\in G\\ &\iff xyx^{-1}=y\text{ for all }x,y\in G\\ &\iff\sigma_x(y)=y\text{ for all }\sigma_x\in \text{Inn}(G), y\in \textbf{G}\\ &\iff \sigma_x = \textbf{id}_{\mathbf{G}}\text{ for all }\sigma_x\in \text{Inn}(G). \end{align*}$$
As I noted in the comments, you are misinterpreting the second part. What it says is: assume that $H$ is normal and abelian, and that $H\not\subset Z(G)$. Then there exists a $g\in G$ such that $\sigma_g|_H$ is not an inner automorphism of $H$; that is, there does not exist an $h\in H$ such that $\sigma_g|_H = \sigma_h$, where $\sigma_h\in\mathrm{Inn}(H)$. Or equivalently, that there does not exist an $h\in H$ such that for every $x\in H$, $gxg^{-1}=hxh^{-1}$.
This is not correctly phrased. What is true is that for every $g\in G$ and for every $h_1\in H$, there exists an $h_2\in H$, which may depend on $h_1$ and on $g$, such that $h_1g=gh_2$. It is not just "some $h_1,h_2$", and the pair may not work for all $g$.
What you know is that because $H$ is normal, then for every $g\in G$ we have $gHg^{-1}=H$. So $\sigma_g(H)=H$. That means that $\sigma_g|_H$ is an automorphism of $H$.
To distinguish the inner automorphisms of $G$ from those of $H$, let us use $\sigma_g\colon G\to G$ to denote the ones that are inner automorphisms of $G$, and let us use $s_h\colon H\to H$ to denote the inner automorphisms of $H$; note that we must have $h\in H$. We know that $\sigma_g|_H\in\mathrm{Aut}(H)$. The question is: will it be in $\mathrm{Inn}(H)$? Does there exist an $h\in H$ such that $\sigma_g|_H=s_h$?
That would require that for every $x\in H$, $$\begin{align*} \sigma_g(x) &= s_h(x)\\ gxg^{-1} &= hxh^{-1}. \end{align*}$$ Now, if $H$ is abelian, then $hxh^{-1}= hh^{-1}x = x$ (or, by the first part, we know that $s_h=\mathrm{id}_H$). So that would require that for every $x\in H$, $gxg^{-1}=x$; or equivalently, that for every $x\in H$, $gx=xg$.
So, I wonder if the fact that $H\not\subset Z(G)$ means that we can find some $g$ for which this does not happen?