order of group element provided coset number I am trying to do an exercise but I am confused at this question. Isn't the order of $a$ just $3$. 
I am trying to use Lagrange Theorem but to no avail since I have no clue what $G$ is. 

 A: $|a|$ clearly divides $30$, for we have:
$a^{30} = (a^3)^{10} = h^{10}$ for some $h \in H$, and by Lagrange, $h^{10} = e$ (since $\langle h\rangle$ is a subgroup of $H$ and its order, which is $|h|$,
 must divide $10$).
On, the other hand, if $|a| = k$, then $a^k = e \in H$, and since $aH$ has order $3$, we must have $3 \mid k$ (for otherwise, $aH = H$, or $a^2H = H$, contradiction).
The possibilities are thus:
$3,6,15,30$.
Letting $G = \Bbb Z_{30}$ and $H = \langle 3\rangle$ (which has order $10$), we see that $1+H$ and $2+H$ both have order $3$ in $G/H$, and:
$10 \in 1+H$ has order $3$ in $G$.
$5 \in 2+H$ has order $6$ in $G$.
$2 \in 2+H$ has order $15$ in $G$, and
$1 \in 1+H$ has order $30$ in $G$, which settles the question.
A: Consider a subgroup $K\subseteq G/H$ generated by $aH$. It follows that $K\simeq G'/H$ for some subgroup $H\subseteq G'\subseteq G$ such that $a\in G'$. Since $|G'/H|=3$ and $|H|=10$ then it follows (by Lagrange's theorem) that $|G'|=30$. And since $a\in G'$ then $|a|$ divides $30$.
Are all divisors of $30$ possible? If you consider $G'=\mathbb{Z}_3\times\mathbb{Z}_{10}$ and $H=0\times\mathbb{Z}_{10}$ then $3, 6, 15, 30$ are clearly possible values of $|a|$. Obviously $1$ is not possible.
What about $2, 5$? (thanks to @DavidWheeler) Of course if $|a|=2$ or $|a|=5$ then $K$ being the image of $\langle a\rangle$ would have to be of order $1,2$ or $5$. But $|K|=3$. Contradiction.
A: A variation on David W's answer.
Write $\langle a\rangle$ for the cyclic subgroup of $G$ generated by $a$.
One has a surjective homomorphism $\pi$
$$ \left< a \right> \to \langle a\rangle H / H,$$
defined by $\pi\colon a^k\mapsto a^k H$. This is a homomorphism, as $H$ is normal. 
The kernel $K$ of $\pi$ is $\langle a\rangle \cap H$. By hypothesis the image of $\pi$ has cardinality $3$. Therefore $\#\langle a \rangle = 3 \ \cdot \#K$. Now, the only possibilities for the cardinality of $K$  are $1,2,5, 10$, as $K$ is a subgroup of $H$. So the only possibilities for the order of $a$ are $3, 6, 15, 30$. By explicit construction (as in the comments, or in David's answer), we see all of these occur.
