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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.

GuPe
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3 Answers3

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

Asinomás
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  • Indeed! It's true however that the cardinality of group of automorphisms of a finite cyclic group must be an even number. If helps the proof somehow. –  Apr 26 '17 at 16:50
  • I realize now I made a mistake, but whether $\text{Aut} (H)$ is cyclic or not is irrelevant, since all we need is that its order is a power of $2$, a condition that does hold. – GuPe Apr 26 '17 at 16:55
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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)

Lukas Heger
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  • But $G/Z(G)\cong \text{Inn}(G)$ (set $\phi (g)= gxg^{-1}$, now $\phi (gh)=\phi(h)\circ \phi(g))$, $\phi$ is also surjective, and by the first isomorphism theorem we have that they are isomorphic because $\text{ker} \phi = Z(G)$. Also, every inner automorphism of $G$ fixes $H$ since $H$ is normal, therefore it is an automorphism of $H$. – GuPe Apr 26 '17 at 17:05
  • @GuachoPerez There's no embedding of $\operatorname{Aut}(H)$ into $\operatorname{Aut}(G)$, in general. – egreg Apr 26 '17 at 17:11
  • @GuachoPerez you have defined a map from $\operatorname{Inn}(G)$ to $\operatorname{Aut}(H)$, but there is no reason for it to be injective, because it may very well be the case that an element acts nontrivially on $G$ by conjugation, but it acts trivially on $H$ by conjugation. – Lukas Heger Apr 26 '17 at 21:40
  • @MatheiBoulomenos so the mistake in the "proof" is saying that $\text{Inn} (G) \subseteq \text{Aut}(H)$ or that $\text{Inn} (G) \cong K \subseteq \text{Aut}(H)$? – GuPe Apr 26 '17 at 21:57
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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.

Nicky Hekster
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