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.

Show that $f_n \rightarrow 0$ in $C([0, 1], \mathbb{R})$

I was given the following problem and was wondering if I was on the right track.

Let $ f_n(x) = \frac{1}{n} \frac{nx}{1 + nx}, \: 0 \le x \le 1$

Show that $ f_n \rightarrow 0$ in $ C([0, 1], \mathbb{R})$ .

I have this theorem that I figured I could use:

$ f_k \rightarrow f$ uniformly on A $ \iff$ $ f_k \rightarrow f$ in $ C_b$ .

In this case, $ C_b$ is the collection of all continuous functions on $ [0,1]$ . So if I can prove the function is uniformly continuous, this would prove that $ f_n \rightarrow 0$ . Can I apply this theorem like this to prove what I want? Also, if I can, would using the Weierstrass M test be best to prove uniform convergence here?