I recently posted this question on math.stackexchange.com. The link is this.
My assertion was “Let be a group with three normal subgroups and such that . Then . This is a generalization of the Third Isomorphism Theorem, which states that , where .”
What was my rationale behind asking this question? Let be a group and its normal subgroup. Then contains elements of the form , where , for every .
Now let be two normal subgroups of such that . Then contains elements of the form and contains elements of the form . Now consider . One coset of this would be all elements of all elements of all elements of all elements of all elements of all elements of . We are effectively adding every element of to all elements of . The most important thing to note here is that every element of is also present in .
Every element of the form any element in in will give the same element in , and by extension in . Let and be two elements in () such that both are not in . Then they will not give the same element in . However, as every element of is individually added to them in , they will give the same element in the latter. If and form different cosets in , then they will also form different cosets in . This led me to conclude that .
This reasoning is however flawed, mainly because need not be a subgroup of . Hence, in spite of heavy intuition into the working of cosets, I got stuck on technicalities.