An attempted generalization of the Third Isomorphism Theorem.

by ayushkhaitan3437

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.