If G has a normal subgroup of index p, prove that G has at least one element of order p. I'm totally lost on this one. 
If $G$ has a normal subgroup of index $p$, prove that $G$ has at least one element of order $p$.
EDIT:
Could you use Cauchy's Theorem?
Let $H$ be a normal subgroup of $G$, with index $p$ where $G$ is finite and $p$ is prime.
Then $(G:H)$ = $|G|$$/$$|H|$ = $p$, by the definition of index.
So $|G| = |H|*p$
By Cauchy's Theorem, if $G$ is a finite group, and $p$ is a prime divisor of $|G|$, then $G$ has an element of order $p$.
 A: The order of $G/N$ is prime, thus it is cyclic.
A: Not-so-overkill argument (compared to Cauchy's theorem):
Suppose $G$ is finite (since, as pointed out by Mathmo123, for $G$ infinite the statement does not hold). Then $G/H$ is of order $p$, hence cyclic, so you can choose a "generator modulo $H$", i.e. $g \in G$ such thet $\langle gH \rangle=G/H$.
Now, order of $g$ is easily seen to be a multiple of $p$. Hence, by taking suitable power of $g$, one obtains an element of order $p$.
A: Take $x\notin N$. Since $G/N\simeq \mathbb{Z}/p\mathbb{Z}$, $x^p\in N$. Now, let $r$ be the order of $x$.
If $p$ doesn't divide $r$, $[x^r]\neq 0$ in $G/N$ because $G/N\simeq\mathbb{Z}/p\mathbb{Z}$ and $[x]\neq 0$, absurd.
Hence $p$ divides $r$: but then, $x^{r/p}\neq 0$ and has order $p$.
A: Hint: There are two other more or less standard proofs that I know of.


*

*Induction on $|G|$. If $|G|=2$ or $3$, it's clear. Otherwise, first prove for abelian $G$ by choosing a nonidentity element and examining its order. Either $|x|$ or $|G/<x>|$ is divisible by $p$; use induction to find an element of order $p$; in the second case, a little work, but not much, is required to turn that into an element of $G$ rather than $|G/<x>|$. In the nonabelian case, if any element has centralizer a multiple of $p$ you are done by induction; otherwise, the class equation tells you that the center is a multiple of $p$ and again you are done by induction.

*A slicker proof starts by examining the group action of $\mathbb{Z}/p$ by rotation on the set $X$ of all $p$-tuples $(g_1, g_2, \dotsc, g_p)\in G^p$ whose product is the identity. By the orbit-stabilizer theorem, every orbit has either size 1 or size $p$, so by divisibility arguments the number of size 1 orbits must be a multiple of $p$. Any element with a size $1$ orbit looks like $(g, g, \dotsc, g)$. But the number of size $1$ orbits cannot be zero, since $(e,e,\dotsc, e)$ has a size 1 orbit. So there must be some other element whose orbit has size 1, i.e. some $g\in G$ such that $g^p = e$.
