## 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.

## Given $f$ is complete and $\lim_{z \rightarrow \infty} \frac{f(z)}{z} = 0$, prove that $f$ is constant

Does anyone know how to solve the following question without zeros poles and Macloren series?

Given $$f$$ is complete and $$\lim_{z \rightarrow \infty} \frac{f(z)}{z} = 0$$, > prove that $$f$$ is constant

Thanks!

## 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?

Thanks