Normal subgroups Recall that if G is a group, X ⊆ G is a subset and g 2 G then we denote gXg−1 := fgxg−1jx 2 Xg. Recall also that if H ≤ G is a subgroup and if g 2 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 2 G and h 2 H we have ghg−1 2 H, that is, if for every g 2 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 2 G we have gHg−1 = H. (3) For every g 2 G we have gH = Hg. Proof. Suppose that (1) holds. Let g 2 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 2 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 2 G be arbitrary. By (3) we have Hg = gH. Therefore for every h 2 H gh 2 Hg, that is, there exists h0 2 H such that gh = h0g. Hence ghg−1 = h0 2 H. Since h 2 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 abelian group then for every H ≤ G we have H/G. (6) For every group G we have G/G and feg /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 center of G. (9) We have V/A4 where V = f, (12)(34); (13)(24); (14)(23)g: (10) Let G = F (a; b) and let H = fw 2 F (a; b)jσa(w) = σb(w) = 0g 1 2 Here for w 2 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 2 G such that g 62 H we have gH 6= H and Hg 6= H. Since [G : H] = 2, it follows that for every g 2 G such that g 62 H we have G=H = fH; gHg and H n G = fH; Hgg, so that G = H t gH = H t Hg: Hence for every g 2 G such that g 62 H we have gH = G − H = Hg. Also, for every g 2 H we have gH = H = Hg. Thus for every g 2 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)i2I is a family such that for every i 2 IHi /G, then \i2I Hi /G. Proposition-Definition 0.6. Let G be a group and let S ⊆ G be a subset of G. Put −1 −1 S^ = [g2GgSg = fgsg jg 2 G; s 2 Sg. Define nclG(S) = hhRii := hS^i ≤ 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 = fa1; : : : ; akg ⊆ G is a finite subset, we denote nclG(a1; : : : ; ak) := nclG(S): Proof. (1) Since S ⊆ S^ ⊆ hS^i = nclG(S), we have S ⊆ nclG(S). We need to show that the subgroup nclG(S) = hS^i ≤ G is normal in G. Note that if s 2 S and a 2 G then (asa−1)−1 = as−1a−1. Therefore S^ [ (S^)−1 = fasa−1js 2 S; a 2 G; 2 f1; −1gg. Let g 2 G and h 2 nclG(S) be arbitrary. By definition of nclG(S) = hS^i it follows that 1 −1 1 −1 h = g1s1 g1 : : : gnsn gn 3 for some n ≥ 0, some s1; : : : ; sn 2 S, g1; : : : ; gn 2 G and 1; : : : ; n 2 {−1; 1g. Then −1 1 −1 2 −1 1 −1 −1 ghg = gg1s1 g1 g2s2 g2 : : : 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 2 hS^i = 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 2 H for every g 2 G, so that S^ ⊆ H. Since H is a subgroup of G, it follows that hS^i ⊆ H. But hS^i = 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 = f(12)g ⊆ Sn. Then nclG(S) = Sn. Recall that every element of Sn can be written as a product of transpositions. There- fore the set T = f(ij)j1 ≤ i < j ≤ ng 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 = fσ (12)σjσ 2 Sng. Thus for S = f(12)g we have S^ = T . Therefore nclG(S) = hS^i = hT i = Sn: Recall that if G is a group and x; y 2 G then the commutator of x and y is [x; y] := x−1y−1xy 2 G. Definition 0.8. Let G be a group. Put S = f[x; y]jx; y 2 Gg. Then the subgroup [G; G] := nclG(S) is called the commutator subgroup of G. Exercise 0.9. Show that if G is a group and S = f[x; y]jx; y 2 Gg then [G; G] = hSi. Hint: Verify that for any x; y; g 2 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)..