(c)Karen E. Smith 2018 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 412. Normal Subgroups DEFINITION: A subgroup N of a group G is normal if for all g 2 G, the left and right N-cosets gN and Ng are the same subsets of G. PROPOSITION: For any subgroup H of a group G, we have jHj = jgHj = jHgj for all g 2 H. THEOREM 8.11: A subgroup N of a group G is normal if and only if for all g 2 G, gNg−1 ⊂ N: Here, the set gNg−1 := fgng−1 j n 2 Ng. NOTATION: If H ⊂ G is any subgroup, then G=H denotes the set of left cosets of H in G. Its elements are sets denoted gH where g 2 G. Recall that the cardinality of G=H is called the index of H in G. A. WARMUP (1) Let 2Z be the subgroup of even integers in Z. Fix any n 2 Z. Describe the left coset n + 2Z (your answer will depend on the parity of n). Describe the right coset 2Z + n. Is 2Z a normal subgroup of Z? What is its index? (2) Let K = h(2 3)i ⊂ S3. Find the right coset K(1 2). Find the left coset (1 2)K: Is K a normal subgroup of S3? (3) Let N = h(1 2 3)i ⊂ S3. Find the right coset N(1 2). Find the left coset (1 2)N. Is gN = Ng for all g 2 G? Is N a normal subgroup of S3? (1) The left coset n + 2Z is the set of odd numbers if n is odd and the set of even numbers if n is even. Ditto for the right coset 2Z + n. The subgroup 2Z is normal because n + 2Z = 2Z + n for all n. Index is two. (2) Right coset K(1 2) is f(1 2); (1 3 2)g. Left coset (1 2)K is f(1 2); (1 2 3)g: Since K(12) 6= (12)K, K is not normal. (3) The right coset N(1 2) = f(12); (2 3); (1 3)g. The left coset is (1 2)N = f(12); (2 3); (1 3)g. We see that (1 2)N = N(1 2). To find the partition into left cosets, we compute all left cosets. The only other coset is eN = fe; (1 2 3); (1 3 2g. We know this because all left cosets have the same cardinality, so since the coset (1 2)N has three element, so do all others. But the cosets are disjoint! So there can be only one more coset, and it is eN. The partition into right cosets is the same! We know that eN = Ne, so one left/right coset is the set fe; (1 2 3); (1 3 2g. The other left/right coset is the complement: N(1 2) = (1 2)N = f(12); (2 3); (1 3)g, which we can also write N(1 3) = (1 3)N = N(2 3) = (3 2)N: So yes, gN = Ng for all g 2 S3. So N is normal. B. EASY PROOFS (1) Prove that feg is a normal subgroup of any group G. (2) Prove that G is a normal subgroup of any group G. (3) Prove that if G is abelian, then every subgroup K is normal. (4) Prove that for any subgroup K, and any g 2 K, we have gK = Kg. (5) Find an example of subgroup H of G which is normal but does not satisfy hg = gh for all h 2 H and all g 2 G. [Hint: There is a six-element G.] (1) For all g 2 G, geg−1 = e 2 feg. So feg is a normal subgroup of G. Alternatively, for all g, the left coset is gfeg equals the right coset fegg; since both are just the one-element set fgg: (2) For all g 2 G, gGg−1 ⊂ G, trivially, since G is closed under composition. So G is normal in G (3) Take arbitrary g 2 G. If G is abelian, we know gK = fgk j k 2 Kg == fkg j k 2 Kg = Kg. So K is normal. (4) If g 2 K, then gk 2 K for all k 2 K, so gK ⊂ K. But also every k 2 K can be written k = g(g−1k) 2 gK, since g−1; k 2 K implies g−1k 2 K. So K = gK. A similar argument shows that K = Kg, so Kg = gK for all g 2 K. (5) The example from A(3) above works: the index two subgroup N = fe; (1 2 3); (1 3 2g is normal. But (1 2)(1 2 3) 6= (1 2 3)(1 2):. C. Let G be the group (Sn; ◦), n ≥ 4. Use Theorem 8.11 to determine which of the following subgroups are 1 normal in Sn. (1) The trivial subgroup e. (2) The whole group Sn. (3) The subgroup An of even permutations. (4) The subgroup H generated by (1 2 3 ). (5) The subgroup Sn−1 of permutations that fix n. (6) Use Lagrange’s Theorem to compute the index of each subgroup in (1)–(5). The trivial and whole subgroups are normal by Problem B. The group An is normal, because given −1 any g 2 Sn, and any h 2 An, we need to check that ghg 2 An. But h is a composition of an even number of transpositions, say 2k transpositions, and if g is a composition of d transpositions, then so is −1 −1 g . So ghg is a composition of d + 2k + d = 2(d + k) transpositions, and hence is in An. The group H = fe; (1 2 3); (1 3 2)g is not normal if n ≥ 4: if we conjugate by (1 5), we get (1 5)(1 2 3)(1 5) = (5 2 3) which is not in H. Similarly, the subgroup Sn−1 of permutations that fix n is not normal: the element (1 2) is in Sn−1 but its conjugate by (1 n) is (2 n) which does not fix n. Lagrange’s theorem tells us that [Sn : An] = 2 (we have even and odd permutations for the cosets) [Sn : H] = n!=3, and [Sn : Sn−1] = n!=(n − 1)! = n. φ D. Let G ! H be a group homomorphism. (1) Prove that the kernel of φ is a normal subgroup of G. × (2) Prove that the map GLn(Q) ! R sending A 7! det A is a group homomorphism. (3) Prove the the group SLn(Q) of determinant one matrices with entries in Q is a normal subgroup of GLn(Q). (1) Take k 2 ker φ and g 2 G arbitrary. We need to show that gkg−1 2 ker φ. Apply φ to get −1 −1 −1 −1 φ(g)φ(k)φ(g ). Since k 2 ker φ, this is φ(g)eH φ(g ) = φ(gg ) = φ(eG) = eH . So gkg 2 ker φ. QED. (2) This is obvious from a Math 217 fact: det A det B = det(AB). × (3) The group SLn(Q) is the kernel of the determinant map GLn(Q) ! Q . So it is normal by (1). 1 −1 It is totally worth using the useful trick you proved on the homework for computing gσg where g and σ are in Sn: for example (1 2 3) ◦ (2 4 6 3)(1 5) ◦ (1 2 3)−1 = [(1 2 3) ◦ (2 4 6 3) ◦ (1 2 3)−1] ◦ [(1 2 3) ◦ (1 5) ◦ (1 2 3)−1] = (3 4 6 1)(2 5)—we got this by applying (1 2 3) to the numbers in the cycle form of the permutation σ = (2 4 6 3)(1 5): E. THE PROOF OF THEOREM 8.11. Let G be a group and H some subgroup. Prove that the following are equivalent: (1) H is normal (2) gHg−1 ⊂ H for all g 2 G, (3) g−1Hg ⊂ H for all g 2 G. (4) g−1Hg = H for all g 2 G. (5) gHg−1 = H for all g 2 G. This is pretty easy, but you should make sure you can do it. The details are in the book; Theorem 8.11, page 251. F. Suppose that H is an index two subgroup of G. (1) Prove that the partition of G up into left cosets is the disjoint union of H and G n H. (2) Prove that the partition of G up into right cosets is the disjoint union of H and G n H. (3) Prove that for every g 2 G, gH = Hg. (4) Prove the THEOREM: Every subgroup of index two in G is normal. (1) If the index is two, there are only two left cosets. Since one is H = eH, the other is its comple- ment (as they are disjoint). (2) Ditto. (3) Take arbitrary g 2 G. There are two cases: g 2 H and g2 = H. If g 2 H, then gH = Hg = H because H is closed under multiplication by g 2 H. If g2 = H, then gH is not the coset H so it must be the coset G n H. But this is also Hg. So gH = Hg. (4) We just saw that gH = Hg for all g. By definition, H is normal. G. OPERATIONS ON COSETS: Let (G; ◦) be a group and let N ⊂ G be a normal subgroup.