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

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}$ don’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.