If $G$ is an abelian group, then $H=\{g\in G \mid |g| \text{ divides }12\}$ is a subgroup of $G$ $$H=\{g\in G \mid |g| \text{ divides }12\}$$
We have to prove that $H$ is a subgroup of $G$. 
Consider $a\in G$ and, the group $\langle a \rangle$ where $|a|=12$. This group will contain elements orders of which will be divide 12. This is a cyclic subgroup I think this should be equal to H but I am not sure as I don't know if G is finite.
 A: Instead of seeking a specific example like cyclic groups, you can prove it in a more general way:
HINT: $H \le G$ if and only if
1) $H$ is not empty.
2) $H$ is closed under binary operation of $G$.
3) $H$ is closed under inverses.
A: Hint: We can rewrite $H=\{g\in G\mid 12g=0\}$.  Therefore, if you can show that the map $g\mapsto 12g$ is a homomorphism, then you have that $H$ is the kernel of a homomorphism and therefore a subgroup.
A: Use the lemma $H$ is a subgroup of $G$ if and only if $\forall a,b \in H : ab^{-1} \in H$ .
Assume $a,b \in H $ then $(ab^{-1})^{12} = a^{12}(b^{12})^{-1} $ and since $o(a)$ and $o(b)$ divide $12$ we have $a^{12}=b^{12}=e $. This implies $(ab^{-1})^{12}=e $ so $o(ab^{-1}) \mid 12 $ and this means $ab^{-1} \in H $.
Note : $a^n=e \Leftrightarrow o(a) \mid n$ .
A: The key ingredient is that if $a,b $ commute, then $|ab|=\rm {lcm}(|a|,|b|) $.  From this we get closure.
The other key ingredient gives us inverses:  $|a^{-1}|=|a|$.
Don't forget to check that $H\ne\emptyset $.  We see that  $e\in H $.
