Completing the proof using the fundamental theorem of cyclic groups $G$ is abelian; $H=${$g\in G,|g| \text { divides } 12$}. Prove that $H$ is a subgroup of $G$.
My idea:  if the order of $g$ divides 12, then the order of the group $G$ must be some multiple of $12$. How can I use the fundamental theorem of cyclic groups to finish this proof?
Thanks in advance!
 A: Hint


*

*$|e|=1$ divides $12$ so $e \in H$.

*Let $a,b \in H$ and let $|a|=m$ and $|b|=n$. Then both $m,n$ divide $12$. Now consider the element $ab$. 


First of all we know that (since $G$ is abelian) $(ab)^{\text{lcm}(m,n)}=e$. Thus $ab$ has a finite order. 
Let $|ab|=k$ then $k$ divides $\text{lcm}(m,n)$. But $\text{lcm}(m,n)$ divides $12$ (common multiple of $m$ and $n$). Therefore $k$ divides $12$ as well. This shows closure.
Now complete the inverse property.
A: 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 $gh \in H$.
