Realizing that $H=\{g\in G : g^{12}=e\}$ simplifies everything because it avoids having to worry about orders.
If you can use homomorphisms, then $H$ is a subgroup because $H=\ker \phi$, where $\phi(x)=x^{12}$, which is a homomorphism $G\to G$ because $G$ is abelian.
Otherwise, it is easy to very directly that $H$ is a subgroup using the characterization above. Again, the key point is that $G$ is abelian. Indeed:
$e\in H$ because $e^{12} = e$.
If $h \in H$, then $(h^{-1})^{12} = (h^{12})^{-1} = e^{-1} =e $ and so $h^{-1} \in H$.
If $g,h \in H$, then $(gh)^{12}=g^{12} h^{12} = e e = e $ and so $gh \in H$.