Quotient groups and isomorphism theorems

Let N be a normal subgroup of G. Define the product of two left cosets by multiplying representatives: the proposed answer for aN times bN is abN. To justify this, replace a by an and b by bm, where n and m belong to N. Their product is ab times the element b inverse n b m, which remains in N by normality and subgroup closure. Thus every choice gives the same resulting coset. Associativity descends from G; the identity is N and the inverse of aN is a inverse N.

Normality is also necessary for this multiplication rule. If it were well defined for a subgroup H, then nH=H for n in H would imply ngH=gH for every g, hence g inverse n g belongs to H. Applying this inclusion for g and g inverse gives conjugation invariance. Without normality, a set of cosets is not a quotient group. Every subgroup defines a partition and an index, but only normal subgroups support this representative-independent multiplication. The projection sending a to aN is a surjective homomorphism with kernel N. For finite G, each coset contains exactly |N| elements, so the quotient has |G| divided by |N| elements.

For a homomorphism f from G to H, the kernel K is normal. Two elements a and b have equal outputs exactly when b inverse a lies in K, equivalently when their cosets agree. Therefore define a map from G/K to the image of f by sending aK to f(a). The fiber calculation proves it is well defined and injective; the definition of image proves it is surjective. It respects multiplication because f does. This proves the first isomorphism theorem, rather than merely matching two group orders.

The theorem describes information, not a convenient coincidence. A quotient removes precisely distinctions invisible to f, while restricting the codomain to the image removes outputs that never occur. Both adjustments matter: G/K need not be isomorphic to the whole stated codomain. In the modular lab, pause, reset, then set both controls to modulus twelve and additive step four. Predict the orbit of zero before playing: its three members are the kernel in the worked example. Construct the other cosets manually by shifting this orbit. The instrument computes one sequence of additive multiples; it does not perform the quotient construction or verify an induced map.

The second isomorphism theorem takes a subgroup H and a normal subgroup N of G. The product HN is a subgroup, N is normal in HN, and H intersect N is normal in H. Restrict the projection to H: its kernel is H intersect N and its image is HN/N. Applying the first theorem gives H/(H intersect N) isomorphic to HN/N. This is a clean way to compare a subgroup before and after collapsing N.

For the third theorem, suppose N is contained in M and both are normal in G. The map from G/N to G/M sending aN to aM is surjective, with kernel M/N. Quotienting in two stages therefore produces the same group as quotienting directly by M. The assumptions are part of the statement: M/N must itself be a normal subgroup of G/N. Always name the actual induced map; doing so makes its domain, kernel, and surjectivity checkable.