Math 402 Assignment 3. Due Wednesday, October 16, 2013 1.In the alternating group A4, let H be the three element cyclic subgroup generated by the element (123). Exhibit the left and right coset of H explicitly. (First tell how many left or right cosets there are.) The number of left or the number of right H-cosets in A4 is ◦(A4)= ◦ (H) = 12=3 = 4. To display the left H-cosets, we will pair H up on the left with a 4-element subgroup K = fI; (12)(34); (13)(24); (14)(23)g of A4: H = IH, (12)(34)H, (13)(24)H, (14)(23)H. To show that none of these cosets duplicate others, we see that none of (12)(34)H, (13)(24)H, (14)(23)H equals H since each contains an element of K not in H. If say (12)(34)H = (13)(24)H, then combining that equation on the left with (13)(24), we get the equation (13)(24)(12)(34)H = H which is not so since (13)(24)(12)(34) = (14)(23) is an element of the left that is not of the right. More generally if H and K are subgroups of G such that H \ K = feg, then the left-H cosets kH, k 2 K, are all distinct. In fact, if kH = k0H for k; k0 2 K, then combining that equation with k0−1 on the left gives the equation k0−1kH = H. Then k0−1k is an element of H by the equation, and it is an element of K since K is a subgroup. Hence, the element must be e, and k = k0. By the same argument, the right cosets of H in G are H = HI, H(12)(34), H(13)(24), H(14)(23). −1 In general, if the left H-cosets are g1H, g2H,....,gnH, then the right H-cosets are Hg1 , −1 −1 Hg2 ,...,Hgn . In particular, if the left coset giH, where gi has order 2, then the right −1 H-coset Hgi equals Hgi. 2. If ◦(G) = pn, where p is prime and n ≥ 1, show that G has an element of order p. Solution. Let a 2 G be any element other than e. Since the order of an element divides the order of the group, and ◦(G) = pn with p prime, we have ◦(a) = pm for some 0 < m ≤ n. Then apm−1 has order p. 3. Let G have order 22. If x 6= e and if y is not in the cyclic subgroup generated by x, show that the subgroup generated by x and y equals G. Solution. For 22, the longest sequences of divisors 1,2,22 and 1,11,22. Hence, by Lagrange's theorem on divisibility, a sequence of subgroups of a group of order 22 has length at most 3. Let C be the cyclic subgroup generated by x and let K be the subgroup generated by x and y. Then C ⊂ K are unequal since y is not an element of H. If K 6= G, then feg ⊂ C ⊂ K ⊂ G 1 would have length 4. Therefore, K = G. 4. Let G be a finite group. (a). Under what circumstances is the map φ : G ! G defined by φ(x) = x2 a homomor- phism? Solution. For φ to be a homomorphism, we need to have the equation φ(ab) = φ(a)φ(b) satisfied for a; b 2 G, i.e., (ab)2 = a2b2, i.e., abab = a2b2. Composing the equation with a−1 on the left and with b−1 on the right, we get the equivalent equation ba = ab. Thus, the \squaring" mapping is a homomorphism if and only if the group G is abelian. (b). Under what circumstance is the map φ an automorphism for an abelian group G? Since G is finite, the mapping is onto if and only if it is 1-1, i.e., if it has kernel equal to feg. The elements in the kernel of the squaring map are just e and the elements of G of order 2. Thus, the mapping is an automorphism if and only if G has no elements of order 2, or equivalently, G has odd order. (We will show that a group of even order has an element of order 2, and more generally, if p is a prime divisor of ◦(G), then G has an element of order p.) Thus, the squaring map is an automorphism when G is an abelian group of odd order. 5. (a). Show that every subgroup H of index 2 in a group G is normal. Solution. Let a be an element of G not in H. Then the two left H coset of G are H and aH and the two right H cosets are H and Ha. Since there are only the two disjoint left cosets, aH = G − H, and similarly, Ha = G − H. Hence, aH = Ha. Thus, the left cosets and the right coset coincide, and so, H is normal. (b). Show by example that a subgroup H of index 3 in a group G need not be normal. In S3, the non-normal subgroup H = fI; (12)g has index 3. H is not normal since the left-H coset (13)fI; (12)g = f(13); (123)g does not equal the right H-coset fI; (12)g(13) = f(13); (321)g. 2.