How can I show all elements of a cyclic subgroup are distinct?
All elements of a finite cyclic group are of the form $\displaystyle a^m$, where $\displaystyle a^n=e$. Imagine that there were two elements $\displaystyle a^m$ and $\displaystyle a^k$ such that k is different from m, but $\displaystyle a^m=a^k$. Let $\displaystyle m=na_1+r_1$ and $\displaystyle k=na_2+r_2$ where the r's are the rest of the division by n and thus less than n. Lets assume, without loss of generality, that $\displaystyle r_1 >r_2$. Then assume $\displaystyle a^m=a^{r_1}$ is equal to $\displaystyle a^{r_2}=a^k$ and we'll get a contradiction. $\displaystyle a^{r_1}=a^{r_2}$ implies $\displaystyle a^{r_1-r_2}=e$ and thus there would be a number $\displaystyle r_1-r_2<n$ such that $\displaystyle a^{r_1-r_2}=e$, which is a contradiction since n is the smallest natural number for which $\displaystyle a^n=e$.
Now do the infinite cyclic one.