### An attempted generalization of the Third Isomorphism Theorem.

I recently posted this question on math.stackexchange.com. The link is this.

My assertion was “Let $G$ be a group with three normal subgroups $K_1,K_2$ and $H$ such that $K_1,K_2\leq H$. Then $(G/H)\cong (G/K_1)/(H/K_2)$. This is a generalization of the Third Isomorphism Theorem, which states that $(G/H)\cong (G/K)/(H/K)$, where $K\leq H$.”

What was my rationale behind asking this question? Let $G$ be a group and $H$ its normal subgroup. Then $G/H$ contains elements of the form $g+H$, where $g+h=(g+\alpha h)+ H$, for every $\alpha\in Z$.

Now let $K_1,K_2$ be two normal subgroups of $G$ such that $K_1,K_2\leq H$. Then $G/K_1$ contains elements of the form $g+K_1$ and $H/K_2$ contains elements of the form $h+K_2$. Now consider $(G/K_1)/(H/K_2)$. One coset of this would be $\{[(g+$ all elements of $K_1)+(h_1+$all elements of $K_2)],[(g+$ all elements of $K_1)+(h_2+$all elements of $K_2)],\dots,[(g+$ all elements of $K_1)+(h_{|H/K_2|}+$all elements of $K_2)]\}$. We are effectively adding every element of $G/K_1$ to all elements of $H$. The most important thing to note here is that every element of $K_1$ is also present in $H$.

Every element of the form $(g+$ any element in $K_1)$ in $G$ will give the same element in $G/K_1$, and by extension in $(G/K_1)/(H/K_2)$. Let $g$ and $g+h$ be two elements in $G$ ($h\in H$) such that both are not in $K_1$. Then they will not give the same element in $G/K_1$. However, as every element of $H$ is individually added to them in $(G/K_1)/(H/K_2)$, they will give the same element in the latter. If $g$ and $g'$ form different cosets in $G/H$, then they will also form different cosets in $(G/K_1)/(H/K_2)$. This led me to conclude that $(G/H)\cong (G/K_1)/(H/K_2)$.

This reasoning is however flawed, mainly because $H/K_2$ need not be a subgroup of $G/K_1$. Hence, in spite of heavy intuition into the working of cosets, I got stuck on technicalities.