Let $A$ be an arbitrary Abelian group and let $H:=\{a\in A\mid\exists b\in A\mid a=b^3\}.$ Prove $H$ is a normal subgroup of $A$. Determine all the possible orders of elements in the group $A/H.$
My thoughts:
I've proven $H$ is a subgroup of $A$ and it is normal since every subgroup of an Abelian group is normal. I then considered an arbitrary element $aH\in A/H, a\in A$ and tried to set some restrictions on its order, i.e., the smallest $k\in\Bbb N$ s. t. $(aH)^k=e_{A/H},$ which boils down to the smallest $k\in\Bbb N$ s. t. $$a^kH=(aH)^k= H\Leftrightarrow a^k\in H\Leftrightarrow\exists b\in A, a^k=b^3.$$ I think that $m_a:=\min\{k\in\Bbb N\mid\exists b\in A, a^k=b^3\}\le 3$ because $3$ is in that set and $a^3=a^3,\forall a\in A.$ I believe the answer would be that $1,2,3$ are all the possible orders, but I'm not sure about that so my question is how to proceed.
For any element $bH$, we have $(bH)^3=H$ i.e. the order divides 3. So, the only possible case will be order 1 and 3.
Regarding the confusion about the existence of elements of the form $x^2=y^3$, note that here $x=\frac{x^3}{y^3}$, so $xH=H$ will have order 1.