# Nonabelian group of order $75$ cannot have a cyclic subgroup of order $25$

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!

• 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. Dec 8 '19 at 15:57

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.
• 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. Dec 9 '19 at 13:51
• @himan $3$ divides the order of $AH$ as does $25$. Dec 9 '19 at 13:57
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 $$G$$.