Is this group always abelian?

Question

Let $|G|$ be odd and its prime factorization contain a prime of the form $r=2^n+1$, if $H$ is a normal subgroup of order $r$, is $G$ abelian?
My reasoning goes like this:
Since $H$ is normal, $G/Z(G)\cong \text{Inn}(G) \subseteq \text{Aut}(H)\cong Z_{2^n}$, therefore $|G/Z(G)|$ must be a power of two by Lagrange's theorem. Because $|G|$ is odd, $Z(G)=G$.
The exercise in the text went like this:
If $G$ has order $3825$ and $H$ is a normal subgroup of order $17$, prove $H\subseteq Z(G)$.
However, this is a weaker result than the one above, which prompts me to believe I might have made a mistake in the generalization, my problem is I do not see where.

Answer 1

the mistake in the proof is assuming $\text{Aut}(H)\cong \mathbb Z_{2^n}$, this must not necessarily be the case.

Answer 2

You don't have $G/Z(G) \subset \operatorname{Aut}(H)$ in general. We do have a natural homomorphism from $G$ to $\operatorname{Aut}(H)$ which sends an element to conjugation by that element, but the kernel of that map is the centralizer of $G$ which may be larger than than $Z(G)$.

(Consider the case that $H$ is the trivial subgroup, for example)

Answer 3

Proposition 1 Let $G$ be a group of odd order and $H$ a normal subgroup with $|H|=p$, where $p$ is a Fermat prime. Then $H \subseteq Z(G)$.

Proof Since $G$ acts by conjugation on $H$ we have an injective homomorphism $G/C_G(H) \hookrightarrow Aut(H)$. Since $H$ is cyclic, $|Aut(H)|=$ power of $2$. But $|G|$ is odd, so $G=C_G(H)$, which is equivalent to $H$ being central.

Proposition 1 (dual) Let $G$ be a group of odd order and $H$ a subgroup with $|G:H|=p$, where $p$ is a Fermat prime. Then $G' \subseteq H$. In particular $H$ is normal.

Proof This is more subtle and uses Burnside's Normal $p$-Complement Theorem, for a proof, see my answer here.