# 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. Sep 28, 2017 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. Sep 28, 2017 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. Sep 28, 2017 at 4:18
• @Prasun That does not seem likely to be helpful since the OP has clearly already considered these subgroups. Sep 28, 2017 at 4:22
• @TobiasKildetoft, Yeah, I just noticed. It would've worked if $G$ was abelian, but then the problem would've become trivial. Sep 28, 2017 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 :-)
– user169852
Sep 28, 2017 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 Sep 28, 2017 at 13:32