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I am working through the following problem, and would like to know if my proof is mostly correct:

Let $G$ be a nonabelian group of order $75$ and $H$ a $5$-Sylow subgroup of $G$. Show that $H$ is not cyclic. (Hint: Show that the conjugation action of $G$ on $H$ is not trivial).

Here is my proof:

Let $\mu:G\times H\to H$ be the group action of conjugation on $H$, i.e. $\mu(g,h)=ghg^{-1}$ for $g\in G$ and $h\in H$. We have that $\ker\mu\unlhd G$, so if $\mu$ is trivial, then this implies that $\ker\mu=G$ and so $G\unlhd G$, which contradicts the fact that $G$ is nonabelian. Note that from $\mu$ we have the associated homomorphism $\phi:G\to\operatorname{Aut}(H)$ and we just showed that $\operatorname{ker}\phi\neq G$.

Let $I$ be a $3$-Sylow subgroup of $G$, and we have the associated homomorphism $\psi:I\to \operatorname{Aut}(H)$ when we let $I$ act on $H$ via conjugation. If $H$ were cyclic, then $H\cong\mathbb Z/25\mathbb Z$ and so $|\operatorname{Aut}(H)|=20$. $|I|$ and $|\operatorname{Aut}(H)|$ are coprime, so $\psi$ must be trivial, implying that $I$ commutes with $H$.

Let $J$ denote the subgroup consisting of $g\in G$ such that $g$ commutes with elements in $H$. Then, $I\leq J$ and so $3$ divides $|J|$ and also $H\leq J$ and so 25 divides $|J|$. $J\leq G$ also, so we get that $|J|=75$ and so $J=C_G(H)=G$. But, from above, we showed that $\operatorname{ker}\phi=C_G(H)\neq G$, so we have a contradiction.

It was not obvious to me how the result followed from the hint, so this is the best thing I could come up with. Thank you for any help or feedback!

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  • $\begingroup$ Why is $\ker(\mu)$ a normal subgroup of $G$? I understand what you mean, but it is wrong written like this. Also $G \lhd G$ is always true, even for nonabelian groups. $\endgroup$ Dec 8 '19 at 15:57
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There are some errors in your proof (which I honestly didn't entirely follow), for example the kernel of $\mu$ is not a subgroup of $G$, but there's one part that solves the whole problem.

Note that if $A$ is a Sylow $3$-subgroup then $AH=G$. Since $H$ is normal, $G$ is a semidirect product $H\rtimes A$. If $A$ acted trivially on $H$ by automorphisms, then $G$ would be abelian because it would be the direct product. As you correctly deduced, if $H$ were cyclic then $A$ would act trivially by order considerations. Thus $H$ is not cyclic.

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  • $\begingroup$ Thank you so much! I think I got stuck on trying to make the hint work, but your answer helped guide me to a correct (and less convoluted) proof. I have one question still: could you please explain why does $A$ being a Sylow $3$-subgroup imply that $AH=G$? Thanks again. $\endgroup$ Dec 9 '19 at 13:51
  • $\begingroup$ @himan $3$ divides the order of $AH$ as does $25$. $\endgroup$ Dec 9 '19 at 13:57
  • $\begingroup$ Oh, okay. Thank you again! $\endgroup$ Dec 9 '19 at 13:59
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Observe from Sylow's theorems that $n_5=1$, so that there is only one Sylow $5$-subgroup and thus it is normal. Let $P$ be that Sylow subgroup. From the $N/C$ theorem, it follows that $N_G(P)/C_G(P) = G/C_G(P)$ is isomorphic to a subgroup of $\mathrm{Aut}(P)$. If $P$ is cyclic then $\mathrm{Aut}(P) \cong C_{20}$. But $P$ is abelian, so $C_G(P) \geq P$, thus the index of $C_G(P)$ in $G$ is either $1$ or $3$. Therefore, if $P$ is cyclic then $P$ is central in $G$.

You should be able to complete this argument.

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