By Nigel Hitchin

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Let G be a group, and N and H subgroups ofG with N normal. Then NH = HN is a subgroup ofG. Proof Since N is a normal subgroup of G, it is easily seen that N H = H N. Now let x,y E NH. We wish to show that xy-l E NH. Since x and yare elements of NH, x = nh and y = nIhl for some elements n,nl EN and h, hI E H. Then, since N is normal, and the result now follows. D. 18] (The Second Isomorphism Theorem). Let G be a group, N a normal subgroup of G and H a subgroup of G. Then N is a normal subgroup of NH, H n N is a normal subgroup of H, and H HN NH (HnN) ~N=N' as illustrated by the following diamond-shaped diagram.

13)' (4). D. 16]. Let R be a ring and [ an ideal of R. The following assertions hold. (1) The quotient set R/ [ is a ring via the operations [a] + [b) = [a + b) and [a] [b] = [a b) . (2) Themap kr : R -+ R/[ defined by kr(x) = [x] (x E R) is a surjective homomorphism of rings with ker (k r) = [. The map kr is called the canonical surjection from R to R/[. Proof. 14)' and is left to the reader. D. Remark. If [ is an ideal of from group theory. (1) If x E R, then [x] = (2) If x, y E R, then, in x -y E I.

15] (The First Isomorphism Theorem). Let I: G _ H be a surjective homomorphism of groups and K = ker (I). Then K is a normal subgroup of G and ~~H G () : K - via the map H defined by ()( [x] ) = I(x) . Proof. 13], K is a normal subgroup of G and, if x, y E G, then in the quotient group GIK, [x] = [y] if, and only if, I(x) = I(y). Hence the map () is both well-defined (that is, independent of the choice of representative of the equivalence class) and injective. The verification that () is an isomorphism of groups is now routine and is left to the reader.