Regarding $|\textrm{G}| = 65$ $\Rightarrow$ $\textrm{G}$ is cyclic.

Question

I am proving that a group of order $\textrm {G}$ being 65 would imply that $\textrm {G}$ is cyclic, using Lagrange's Theorem and $N \textrm{by} C$ Theorem.

Clearly, one can show that not all non-identity elements in $\textrm{G}$ would have order 13. But I have difficulty in proving that not all non-identity elements in $\textrm{G}$ would have order 5 either.

Assuming that all non-identity elements in $\textrm{G}$ have order 5, the following statement can be concluded:

1) Every non-trivial, proper subgroup of $\textrm{G}$ is a cyclic group of order 5.

2) None of those groups will be normal subgroups of $\textrm{G}$; otherwise it will imply that $\textrm{G}$ will have a subgroup of order 25 which is not possible since $|\textrm{G}| = 65$.

3) Index of every non-trivial, proper subgroup of $\textrm{G}$ will be 13.

My hunch is that point (3) would possibly show that $\textrm{G}$ doesn't have every non-identity element of order 5. But I have difficulty in showing that given any non-trivial, proper subgroup, I can construct more than 13 coset of that subgroup. Or is their any other fact that I'm missing.

Answer 1

General Result:

Let $G$ be a group and $\vert G\vert = pq$, where $p$ and $q$ are distinct primes with $p < q$ and $p \nmid q-1$. Then $G$ is cyclic.

Proof. Let $n_p$ be the number of Sylow $p$-subgroups of $G$. Then, by Sylow's Theorem, $n_p \mid q$ and $n_p = 1\pmod p$. This implies that either $n_p = 1$ or $n_p = q$. Firstly, suppose that $n_p = q$. Then $q = 1\pmod p$, and so $p \mid q-1$ - a contradiction!

Thus $n_p = 1$, and so if $P$ is a Sylow $p$-subgroup of $G$, then $P$ is unique. Thus $P \trianglelefteq G$, and by the same token $Q \trianglelefteq G$, where $Q$ is the unique Sylow $q$-subgroup of $G$. Because $\vert P\vert = p$ and $\vert Q\vert = q$, where $p$ and $q$ are primes, we can write $P = \langle x \rangle$ and $Q = \langle y \rangle$ for some $x,y\in G$ with $o(x) = p$ and $o(y)=q$. Also, since $p$ and $q$ are distinct, $P \cap Q = \lbrace 1\rbrace$ by Lagrange's Theorem. Now the commutator $[x,y] = x^{-1}y^{-1}xy \in P\cap Q = \lbrace 1 \rbrace$ (because $P$ and $Q$ are normal in $G$). Thus $x^{-1}y^{-1}xy = 1$, i.e. $xy = yx$. Then it follows easily that $o(xy) = pq$, and so $G = \langle xy \rangle$ is cyclic.

In your example, $\vert G \vert = 65 = 5\cdot 13$, so choose $p = 5$ and $q = 13$. Clearly $5 < 13$ and $5 \nmid 12$ so the hypothesis of the above result are satisfied. Then $G$ is cyclic.

Hope this helps.

For further information on Sylow's Theorems: https://en.wikipedia.org/wiki/Sylow_theorems