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Home , Normal subgroup

Normal

Recall that if G is a , X ⊆ G is a and g ∈ G then we denote gXg−1 := {gxg−1|x ∈ X}. Recall also that if H ≤ G is a and if g ∈ G then gHg−1 is again a subgroup of G, called the conjugate of H by g. Definition 0.1. Let G be a group and let H ≤ G be a subgroup. We say that the subgroup H is normal in G, denoted H/G, if for every g ∈ G and h ∈ H we have ghg−1 ∈ H, that is, if for every g ∈ H we have gHg−1 ⊆ H. Theorem 0.2. Let G be a group and let H ≤ G be a subgroup. Then the following conditions are equivalent: (1) We have H/G. (2) For every g ∈ G we have gHg−1 = H. (3) For every g ∈ G we have gH = Hg. Proof. Suppose that (1) holds. Let g ∈ G be arbitrary. Since H is normal in G and since (g−1)−1 = g we have gHg−1 ⊆ H and g−1Hg ⊆ H. By multiplying g−1Hg ⊆ H by g on the left and by g−1 on the right we get H ⊆ gHg−1. Since we also have gHg−1 ⊆ H, it follows that gHg−1 = H, as required. Thus (1) implies (2). Suppose now that (2) holds. Let g ∈ G be arbitrary. Then gHg−1 = H. By multiplying this equality on the right by g we get gH = Hg, as required. Thus (2) implies (3). Suppose now that (3) holds. Let g ∈ G be arbitrary. By (3) we have Hg = gH. Therefore for every h ∈ H gh ∈ Hg, that is, there exists h0 ∈ H such that gh = h0g. Hence ghg−1 = h0 ∈ H. Since h ∈ H was arbitrary, this implies that gHg−1 ⊆ H. Therefore H/G. Thus (3) implies (1).  Example 0.3.

(1) For every n ≥ 1 we have An /Sn. (2) We have SL(2, R) / GL(2, R). (3) We have SL(2, Z) /SL6 (2, R). (4) We have h(12)i/S 6 3. (5) If G is an then for every H ≤ G we have H/G. (6) For every group G we have G/G and {e} /G. (7) We have F (a, b) /F6 (a, b, c). (8) For every group G we have Z(G) /G, where Z(G) is the of G. (9) We have V/A4 where V = {, (12)(34), (13)(24), (14)(23)}. (10) Let G = F (a, b) and let

H = {w ∈ F (a, b)|σa(w) = σb(w) = 0} 1 2

Here for w ∈ F (a, b) σa(w) is the exponent sum on a in w, and σb(w) is the −1 −1 −1 −1 exponent sum on b in w. E.g. σa(a bbbab ) = 0 and σb(a bbbab ) = 2. Then H/F (a, b). Theorem 0.4. Let G be a group and let H ≤ G be a subgroup of G such that [G : H] = 2. Then H/G. Proof. Note that for every g ∈ G such that g 6∈ H we have gH 6= H and Hg 6= H. Since [G : H] = 2, it follows that for every g ∈ G such that g 6∈ H we have G/H = {H, gH} and H \ G = {H, Hg}, so that G = H t gH = H t Hg. Hence for every g ∈ G such that g 6∈ H we have gH = G − H = Hg. Also, for every g ∈ H we have gH = H = Hg. Thus for every g ∈ G we have gH = Hg. Therefore by Theorem 0.2 we have H/G.  Proposition 0.5. Let G be a group. (1) If H/G and K/G then H ∩ K/G. (2) If (Hi)i∈I is a family such that for every i ∈ IHi /G, then ∩i∈I Hi /G. Proposition-Definition 0.6. Let G be a group and let S ⊆ G be a subset of G. Put −1 −1 Sˆ = ∪g∈GgSg = {gsg |g ∈ G, s ∈ S}. Define

nclG(S) = hhRii := hSˆi ≤ G. Then

(1) We have nclG(S) /G and S ⊆ nclG(S). (2)

nclG(S) = ∩ H, S⊆H/G

so that nclG(S) is the intersection of all normal subgroups of G containing S. The subgroup nclG(S) is called the normal closure of S in G, or the subgroup normally generated by S. If S = {a1, . . . , ak} ⊆ G is a finite subset, we denote

nclG(a1, . . . , ak) := nclG(S).

Proof. (1) Since S ⊆ Sˆ ⊆ hSˆi = nclG(S), we have S ⊆ nclG(S). We need to show that the subgroup nclG(S) = hSˆi ≤ G is normal in G. Note that if s ∈ S and a ∈ G then (asa−1)−1 = as−1a−1. Therefore Sˆ ∪ (Sˆ)−1 = {asa−1|s ∈ S, a ∈ G,  ∈ {1, −1}}. Let g ∈ G and h ∈ nclG(S) be arbitrary. By definition of nclG(S) = hSˆi it follows that

1 −1 1 −1 h = g1s1 g1 . . . gnsn gn 3 for some n ≥ 0, some s1, . . . , sn ∈ S, g1, . . . , gn ∈ G and 1, . . . , n ∈ {−1, 1}. Then

−1 1 −1 2 −1 1 −1 −1 ghg = gg1s1 g1 g2s2 . . . gnsn gn g = 1 −1 −1 2 −1 −1 −1 1 −1 −1 gg1s1 g1 g gg2s2 g2 g g . . . g ggnsn gn g = 1 −1 2 −1 1 −1 (gg1)s1 (gg1) (gg2)s2 (gg2) ... (ggn)sn (ggn) −1 Therefore ghg ∈ hSˆi = nclG(S), and hence nclG(S) /G, as required. (2) We already know by part (1) that there exists at least one normal subgroup of G containing S, namely nclG(S). Denote A := ∩ H. Therefore by Proposition 0.5, S⊆H/G S ⊆ A/G. Since nclG(S) is a normal subgroup of G containing S, from the definition of A we −1 have A ⊆ nclG(S). On the other hand, if S ⊆ H/G, then gsg ∈ H for every g ∈ G, so that Sˆ ⊆ H. Since H is a subgroup of G, it follows that hSˆi ⊆ H. But hSˆi = nclG(S), so that nclG(S) ⊆ H. Thus for every H such that S ⊆ H/G we have nclG(S) ⊆ H. Therefore nclG(S) ⊆ ∩ H = A. Since we already know that A ⊆ nclG(S), it follows that nclG(S) = A, S⊆H/G and part (2) is verified. 

Example 0.7. Let n ≥ 2, G = Sn and S = {(12)} ⊆ Sn. Then nclG(S) = Sn. Recall that every element of Sn can be written as a product of transpositions. There- fore the set T = {(ij)|1 ≤ i < j ≤ n} of all transpositions in Sn generates Sn as a subgroup, that is hT i = Sn. We also know that any two transpositions in Sn are conjugate, so that −1 T = {σ (12)σ|σ ∈ Sn}. Thus for S = {(12)} we have Sˆ = T . Therefore nclG(S) = hSˆi = hT i = Sn. Recall that if G is a group and x, y ∈ G then the of x and y is [x, y] := x−1y−1xy ∈ G. Definition 0.8. Let G be a group. Put S = {[x, y]|x, y ∈ G}. Then the subgroup [G, G] := nclG(S) is called the of G. Exercise 0.9. Show that if G is a group and S = {[x, y]|x, y ∈ G} then [G, G] = hSi. Hint: Verify that for any x, y, g ∈ G g[x, y]g−1 = [gxg−1, gyg−1] and use this fact to show that hSi /G. Example 0.10. Let G = F (a, b) and let H/F (a, b) be the subgroup from part (10) of Example 0.3. Then H = [F (a, b),F (a, b)] = nclF (a,b)([a, b]). (Try to prove these two equalities – both are fairly hard).