Let $G$ be a finite commutative group and $H, K$ be two subgroups (hence normal) of $G$ such that for some some $g\in G$, $H=\langle g\rangle$, $K=\langle g^2 \rangle$ and $|g|=|g^2|$.
Can we say the factor groups $G/H$ and $G/K$ are isomorphic? If yes, can we further tell if they are equal? And if they are non-isomorphic, what should I add to establish isomorphism or equality?
I believe the factor groups $G/H$ and $G/K$ are not only isomorphic, they are equal as well. Both factor groups are in the same order. I tried to show they are equal, by choosing an arbitrary element $\alpha:=aH$ from $G/H$. This would be equal to $bK\in G/K$ for some $b\in G$, provided $$aH=bK\Leftrightarrow a\langle g\rangle=b\langle g^2 \rangle$$ is solvable for $b$. The problem now reduces to determine $b$.
To establish this, let $x\in a\langle g\rangle$. Then $x=ag^r$ for some integer $r$ and so $g^r=a^{-1}x$ i.e. $g^{2r}=a^{-2}x^2$, since $G$ is commutative. Then $x=x^{-1}a^2 (g^2)^r$. Choosing $b=x^{-1}a^2$ we finally have $x\in b\langle g^2\rangle$. Thus $aH\subseteq bK$. Proceeding, in the same manner, $bK\subseteq aH$ and finally $\alpha\in G/K$.
Is this correct? Any kind of help would be highly appreciated. Also, feel free to edit if any correction is required in typesetting.