Finding a subgroup $MN$ of $\Bbb Z_{120}$ given $N$ and $M.$ Is the following question I found in a past exam paper a trick question ?

$$G=\Bbb Z_{120}, M=\{0,10,20,...,110\}, N=\{0,4,8,12,..,116\}.$$ Then identify the subgroup of $G$ given by $MN$. 

The reason I think it's a trick question is that $|M|=11,|N|=29$ so wouldn't this imply that $|MN|=319$ and then by Lagrange can't be a subgroup as the order of this group doesn't divide G ? 
 A: It's more customary to write $M+N$ when the group is written with additive notation.
First you're wrong in $|M|$ and $|N|$, which are $12$ and $30$ respectively. On the other hand, $|M+N|$ cannot be $360$, because you only have $120$ elements to choose from.
The general formula, holding also for nonabelian groups, is
$$
|M+N|=\frac{|M|\,|N|}{|M\cap N|}
$$
The formula is easier to prove for abelian groups. Still using additive notation, consider the surjective group homomorphism
$$
\sigma\colon M\times N\to M+N,\qquad (x,y)\mapsto x-y
$$
Its kernel is the set of pairs $(x,y)$ such that $x=y$, which is in an obvious bijection with $M\cap N$. The homomorphism theorem says that
$$
(M\times N)/\!\ker\sigma\cong M+N
$$
so that
$$
|M+N|=\frac{|M\times N|}{\lvert\ker\sigma\rvert}=\frac{|M|\,|N|}{|M\cap N|}
$$
Can you tell what's $M\cap N$?
A different way is to notice that $M=10\mathbb{Z}/120\mathbb{Z}$ and $N=4\mathbb{Z}/120\mathbb{Z}$, so
$$
M+N=(10\mathbb{Z}+4\mathbb{Z})/120\mathbb{Z}=2\mathbb{Z}/120\mathbb{Z}
$$
because $2=\gcd(10,4)$.
A: It's not a trick question! Looks like you made an off-by-one error - it happens to all of us. In particular, you wrote $|M|=11$, presumably because the last value is $11 \times 10$ and you're counting up in $10$s. But let's write out the elements of $M$ fully and see what happens:
$$M= \{0,10,20,30,40,50,60,70,80,90,100,110\}$$
Counting them by hand gives $|M|=12$, and similarly $|N|=30$, as we'd expect for subgroups of $G$. 
For a simpler example of this kind of mistake, notice that if we wanted the size of $A = \{0,1,2,3\}$, you might guess it's $3$ because the highest value is $3$, but there are actually four elements - since $0$ is included.
I should also point out that $|M|=12$, $|N|=30$ doesn't imply that $|MN|=360$, see if you can spot why.
