Discrete Mathematics

Normal subgroups: A normal subgroup is a special kind of subgroup of a group. Recall from the last chapter taht any subgroup H has right and left cosets, which may not be the same. We say that H is a normal subgroup of G if the right and left cosets of H in G are the same; that is, if Hx = xH for any x 2 G.
There are several equivalent ways of saying the same thing. We define
x−1Hx = {x−1hx : h ∈H}
for any element x ∈ G.

Proposition 3.19 Let H be a subgroup of G. Then the following are equivalent:
(a) H is a normal subgroup, that is, Hx = xH for all x ∈ G;
(b) x−1Hx = H for all x ∈ G;
(c) x−1hx 2 H, for all x ∈ G and h ∈ H.

Proof If Hx = xH, then x−1Hx = x−1xH = H, and conversely. So (a) and (b) are equivalent. If (b) holds then every element x−1hx belongs to x−1Hx, and so to H, so (c)
holds. Conversely, suppose that (c) holds. Then every element of x−1Hx belongs to H, and we have to prove the reverse inclusion. So take h ∈ H. Putting y = x−1, we have k = y−1hy = xhx−1 ∈ H, so h ∈ x−1Hx, finishing the proof. Now the important thing about normal subgroups is that, like ideals, they are kernels of homomorphisms.

Proposition 3.20 Let q : G1->G2 be a homomorphism. Then Ker(q) is a normal subgroup of G1.
Proof Let H = Ker(q). Suppose that h 2 H and x ∈ G. Then
(x−1hx)θ = (x−1)θ · hθ · xθ = (xθ)−1 · 1 · xθ = 1,
so x−1hx ∈ ker(θ) = H. By part (c) of the preceding Proposition, H is a normal subgroup of G.
There are a couple of situations in which we can guarantee that a subgroup is normal.

Proposition 3.21 (a) If G is Abelian, then every subgroup H of G is normal. (b) If H has index 2 in G, then H is normal in G.
Proof (a) If G is Abelian, then xH = Hx for all x ∈ G.
(b) Recall that this means that H has exactly two cosets (left or right) in G.

One of these cosets is H itself; the other must consist of all the other elements of G, that is, G\H. This is the case whether we are looking at left or right cosets. So the left and right cosets are the same. Remark We saw in the last chapter an example of a group S3 with a non-normal subgroup having index 3 (that is, just three cosets). So we can’t improve this theorem from 2 to 3.

In our example in the last section, the subgroup {1, (1,2,3), (1,3,2)} of S3 has index 2, and so is normal, in S3; this also follows from the fact that it is the kernel
of a homomorphism. For the record, here is a normal subgroup test:

Proposition 3.22 (Normal subgroup test) A non-empty subset H of a group G is a normal subgroup of G if the following hold:

(a) for any h,k ∈ H, we have hk−1 ∈ H;
(b) for any h ∈ H and x ∈ G, we have x−1hx ∈ H.
Proof (a) is the condition of the Second Subgroup Test, and we saw that (b) is a condition for a subgroup to be normal.