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**Deveno** Incidentally, the reason why we want normal subgroups is so that coset multiplication makes sense.

We would hope that (Na)(Nb) = N(ab).

If Ng = gN for all elements g of G, then a typical element of **(Na)(Nb) = {nan'b: n,n' in N} is nan'b, and since aN = Na, an' = n''a, so that nan'b = (nn'')ab, which is in N(ab).**

On the other hand we have for any element nab in N(ab): nab = an'b = (ea)(n'b), which is in (Na)(Nb).

If, however, Ng does not equal gN, for some particular g, the above argument doesn't work.

As johng has indicated, S_{3} provides an example: let N = {e, (1 2)}, and consider N(2 3) = {(2 3), (1 2 3)}, and N(1 3) = {(1 3), (1 3 2)}.

We would hope that N(2 3)N(1 3) = N((2 3)(1 3)) = N(1 2 3).

First, we calculate N(1 2 3) = {(1 2 3), (2 3)}.

Next, we calculate N(2 3)N(1 3):

(2 3)(1 3) = (1 2 3), so far, so good.

(2 3)(1 3 2) = (1 2) <---this is a problem, it isn't in N(1 2 3)

(1 2 3)(1 3) = (2 3), this would have been OK

(1 2 3)(1 3 2) = e, another problem.

As you can see, N(2 3)N(1 3) has 4 distinct elements, so it cannot possibly be ANY coset of N (all of which have just 2 elements).

So for that subgroup, there is no feasible way to define "coset multiplication".