index of a group and element of a center order are co-prime This is a weird question:

Let $G$ be a group and $H$ its subgroup of index 5. $a \in Z(G)$ and $ord(a)=3$. Prove that $a\in H$.

So my idea was to look at $\langle aH\rangle$, we know that $H \le \langle aH\rangle$  and that:
$5=[G:H] = [G: \langle aH\rangle]\cdot [\langle aH\rangle:H]$
therefore since 5 is a prime we have the two situations:
i) $[G:\langle aH\rangle]=5$ and $[\langle aH\rangle:H]=1$ - in which $H=\langle aH\rangle$ thus $a\in H$
ii) $[G:\langle aH\rangle]=1$ and $[\langle aH\rangle:H]=5$ - in which we get $G=\langle aH\rangle$ but I fail to see a contradition here.
Will appreciate any help, thanks!

Indeed, you are very close. You already know that $[\langle aH\rangle : H] \mid 5$. 
Suppose for the sake of contradiction that $a \notin H$. Then the elements of $\langle aH\rangle$ are of the form $a^kh$ for $h\in H$. Since the order of $a$ is $3$, this means that the distinct cosets are $\{H, aH, a^2H\}$, so that $[\langle aH\rangle : H] = 3$, which contradicts the fact that $[\langle aH\rangle : H] \mid 5$.
