# Group of order $2555$ is cyclic

I am preparing for an exam and am stuck on the following problem :

Let $G$ be a group of order $2555 = 5 \cdot 7 \cdot 73$, show that $G$ is cyclic.

It is not hard to show that the Sylow-73 subgroup is normal in $G$ and that at either the Sylow-5 or the Sylow-7 subgroup is normal. My thought was to prove that both the the Sylow-5 and Sylow-7 subgroups are normal, because then the claim would follow, but I am unsure how to proceed.

Perhaps using the fact that $G$ mod the Sylow-73 subgroup is cyclic? Any hint would be appreciated.

EDIT: So it suffices to show that $G$ is abelian because there is only one abelian group of order 2555, namely the cyclic group of order $2555$. Let $P_{73}$ denote the Sylow 73-subgroup. It is easy to see that $G/P_{73}$ is cyclic. Further we have that $G$ is abelian if $P_{73} \subseteq Z(G)$.

I believe that $P_{73} \subseteq Z(G)$ can (may?) be shown by letting $G$ act on $P_{73}$ via conjugation. Then by the class equation, $73 = \sum_{i =1}^n \text{cl}(x_{i})$ where $x_{1}, \dots, x_{n}$ is a set of representatives of each conjugacy class. I think that one can show that $\text{cl}(x_{i})$ is trivial, which implies that $gx_{i}g^{-1} = x_{i}$ for all $g \in G$ or, equivalently, that $gx_{i} = x_{i}g$. This would show that each element of $P_{73}$ is an element of the center of $G$ and then the claim would follow.

• Would you show us what you've tried? Showing that all of the subgroups are normal would be useful, especially in the context of having finite cyclic groups classified. – Chickenmancer Sep 28 '17 at 4:10
• @Chickenmancer I have outlined pretty much everything I have done so far. I didn't think it was necessary to fill in all the details. – Oiler Sep 28 '17 at 4:17
• Hint: All the factors $5,7,73$ are primes. Apply Cauchy's theorem to show that an element exists whose order equals the group order. – Prasun Biswas Sep 28 '17 at 4:18
• @Prasun That does not seem likely to be helpful since the OP has clearly already considered these subgroups. – Tobias Kildetoft Sep 28 '17 at 4:22
• @TobiasKildetoft, Yeah, I just noticed. It would've worked if $G$ was abelian, but then the problem would've become trivial. – Prasun Biswas Sep 28 '17 at 4:24

I'm pretty sure this should do it, let me know if there are any mistakes. As you noticed, it's not hard to show that the group of order $73$ is normal. Let $H$ be that group and let $K$ be the group of order $5 \cdot 7$. Then by order considerations, $H \cap K = \{e\}$ and $HK=G$. Hence $G$ is the semi direct product of $H$ and $K$, and we have a map $\varphi: K \to Aut(H)$. But $Aut(H)$ is a group of order $72$, and since $\varphi(K)$ divides $K$, $\varphi$ must be trivial.

Hence $G \cong \mathbb{Z}/73\mathbb{Z} \times K$. By similar reasoning, one can show that $K \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/7 \mathbb{Z}$. Hence $G \cong \mathbb{Z}/73\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/7 \mathbb{Z}$.

Let $H$ be a subgroup of order $5$ and $K$ be a subgroup of order $7$. As you already showed, Sylow counting forces at least one of $H$ or $K$ to be normal. Therefore $L = HK$ is a subgroup. Its order is $|L| = |HK| = |H||K|/|H \cap K| = 35$.

Let $P$ denote the unique subgroup of order $73$, and observe that we have $G = LP$, which is a semidirect product since $P \lhd G$ and $L \cap P = 1$.

Now, since $P$ is cyclic with prime order $73$, we have that $\operatorname{Aut}(P)$ has order $72$. Let $\phi : L \to \operatorname{Aut}(P)$ denote the homomorphism such that $\phi(x)$ is elementwise conjugation of $P$ by $x$ (this is indeed an automorphism of $P$ since $P$ is normal). The order of the image of this homomorphism, $|\operatorname{im}(\phi)|$, must divide both $|L| = 35$ and $|\operatorname{Aut}(P)| = 72$, hence $|\operatorname{im}(\phi)| = 1$, which means that $\phi$ is the trivial map. Thus each $x \in L$ conjugates each element of $P$ trivially, so $LP$ is in fact a direct product.

In particular, $L \lhd G$. As $H$ and $K$ are both characteristic subgroups of $L$, it follows that $H$ and $K$ are both normal in $G$, from which we can conclude that $L = H \times K$ and thus $G = H \times K \times P$. This forces $G$ to be abelian, and then it easily follows by order considerations that $G$ must be cyclic.

• Oops, I see that @leibnewtz answered while I was composing this, with more or less the same solution. I'll leave mine here in case the slight extra detail is of any use :-) – Bungo Sep 28 '17 at 6:40
• I had to go with @leibnewtz answer because he beat you to it, but I appreciated the details you filled in nonetheless – Oiler Sep 28 '17 at 13:32