Abstract Algebra -- isomorphism question If N, M are normal subgroups of G, prove that NM/M is isomorphic to N/N intersect M. That's how the problem reads, although I am not sure how to make the proper upside-down cup intersection symbol appear on this forum. Or how to make the curly "=" sign for an isomorphism. First of all, I am uncertain whether in fact N/N intersect M is meant to read N/(N intersect M), but I am going on that assumption. Otherwise, N/N would just be N, wouldn't it? The collection of groups given by NM/M can be found by taking group M and multiplying by members of N and M: Mnm. The groups are normal, so this can be rewritten as nMm. The expression Mm = M since a group multiplied by one of its own members is itself. If we wish, we can move the n so that we end up with a simple right coset: Mn. Basically then it seems to me that NM/M = the collection of right cosets Mn, where n is an element of N. If n is in the intersection of M and N then Mn = M. However, I am uncertain how to proceed if n is not in the intersection of M and N. Perhaps that does not require simplification. In any case, I am pretty mystified on how to proceed with the other expression of interest, N/N intersect M.