# 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.