2010 Spring MATH 4320: Solutions to Prelim 2 Instructor: Yuri Berest Problem 1. (a) This is straightforward: for example, we have 1 [(x1 , x2 ) (y1 , y2 ))] = 1 [(x1 y1 , x2 y2 )] = x1 y1 = 1 [(x1 , x2 )] 1 [(y1 , y2 )] , and similarly for 2 . Both 1 and 2 are surjective, because given any x1 H1 and any x2 H2 , we have 1 [(x1 , x2 )] = x1 and 2 [(x1 , x2 )] = x2 . Finally, Ker(1 ) = (e1 , H2 ) and Ker(2 ) = (H1 , e2 ) . (b) Given any group G with two homomorphisms f1 : G H1 and f2 : G H2 , we dene : G H1 H2 by (x) := (f1 (x), f2 (x)) for x G. Then, we see at once that (1 )(x) = 1 [(x)] = 1 [(f1 (x), f2 (x))] = f1 (x) and (2 )(x) = 2 [(x)] = 2 [(f1 (x), f2 (x))] = f2 (x) for all x G . In other words, f1 = 1 and f2 = 2 as required. Now, suppose there is another homomorphism, say : G H1 H2 , satisfying f1 = 1 and f2 = 2 . By denition of H1 H2 , the elements (x) in the image of can be written as (x) = (g1 (x), g2 (x)) , where g1 (x) H1 and g2 (x) H2 . Since f1 = 1 and f2 = 2 , we have f1 (x) = (1 )(x) = 1 [ (x)] = 1 [(g1 (x), g2 (x))] = g1 (x) and f2 (x) = (2 )(x) = 2 [ (x)] = 2 [(g1 (x), g2 (x))] = g2 (x) for all x G. Thus, g1 (x) = f1 (x) and g2 (x) = f2 (x) , and therefore (x) = (x) for all x G. This proves that = , which means the uniqueness of . Problem 2. (a) For each h H and for all x1 and x2 in N , we have f (h)(x1 x2 ) := h (x1 x2 ) h1 = h x1 (h1 h) x2 h1 = (h x1 h1 ) (h x2 h1 ) . So f (h)(x1 x2 ) = f (h)(x1 ) f (h)(x2 ) , which means that f (h) is a group homomorphism N N . Moreover, f (h) is bijective for each h H , because it has an inverse (namely f (h1 )). Thus, f : H Aut(N ) is a well-dened map, assigning to elements of H the group automorphisms of N . It remains to show that this map is a homomorphism of groups.
Fix any h1 and h2 in H . Then f (h1 h2 ) : N N is given by f (h1 h2 )(x) := (h1 h2 ) x (h1 h2 )1 = (h1 h2 ) x (h1 h1 ) = h1 (h2 x h1 ) h1 , 1 2 1 2 where x N . Since h2 x h1 = Adh2 (x) = f (h2 )(x) , we see that 2 f (h1 h2 )(x) = h1 [f (h2 )(x)] h1 = f (h1 )[f (h2 )(x)] = [f (h1 ) f (h2 )](x) . 1 This holds for all x N , so we conclude f (h1 h2 ) = f (h1 )f (h2 ) as functions N N . Since the composition is precisely the group operation on Aut(N ), the last equality implies that f is a group homomorphism. (b) The map : N H N H , (x, h) xh , is obviously surjective. To show that is injective consider (x1 , h1 ) and (x2 , h2 ) in N H , such that x1 h1 = x2 h2 . The last equation is equivalent to x1 x2 = h1 h1 in 1 2 G . Now, since x1 x2 N , while h1 h1 H , we must have x1 x2 = e 1 2 1 and h1 h1 = e , because N H = {e} . It follows that x1 = x2 in N and 2 h1 = h2 in H , and therefore (x1 , h1 ) = (x2 , h2 ) in N H . The map is an isomorphism of groups i [(x1 , h1 ) (x2 , h2 )] = [(x1 x2 , h1 h2 )] = [(x1 , h1 )] [(x2 , h2 )] . for all (x1 , h1 ) and (x2 , h2 ) in N H . This last equation says that x1 x2 h1 h2 = x1 h1 x2 h2 (1)
for all x1 , x2 N and h1 , h2 H . Letting x1 = h2 = e in (1), we see that x2 h1 = h1 x2 for all x2 N and h1 H . Conversely, if x2 h1 = h1 x2 for all x2 N and for all h1 H , then (1) obviously holds. Thus, being a group homomorphism is equivalent to the condition x h = h x for all x N and for all h H . The latter can be written as h x h1 = x , or equivalently f (h)(x) = x . The last equation simply says that the map f sends every element h H to the identity map IdN on N . (c) It is routine to check that (G, ) satises the axioms of a group. For example, lets check the associativity of : [(x1 , h1 ) (x2 , h2 )] (x3 , h3 ) = (x1 h1 (x2 ), h1 h2 ) (x3 , h3 ) = ((x1 h1 (x2 )) h1 h2 (x3 ), (h1 h2 ) h3 ) = (x1 h1 (x2 ) (h1 h2 )(x3 ), h1 h2 h3 ) .
On other the hand, (x1 , h1 ) [(x2 , h2 ) (x3 , h3 )] = (x1 , h1 ) (x2 h2 (x3 ), h2 h3 ) = (x1 h1 (x2 h2 (x3 )), h1 (h2 h3 )) = (x1 h1 (x2 ) h1 (h2 (x3 )), h1 h2 h3 ) = (x1 h1 (x2 ) (h1 h2 )(x3 ), h1 h2 h3 ) . Comparing the expressions in the right-hand sides, we conclude that [(x1 , h1 ) (x2 , h2 )] (x3 , h3 ) = (x1 , h1 ) [(x2 , h2 ) (x3 , h3 )] . Note, in the above calculations we used the formulas h1 h2 (x) = (h1 h2 )(x) and h (x y ) = h (x) h (y ) , which follow from the fact that : H Aut(N ) is a group homomorphism. The maps N G , x (x, eH ) , and H G , h (eN , h) , are obviously injective. Lets check that these are group homomorphisms. For example, under the rst map the product x1 x2 N goes to (x1 x2 , eH ), while by denition of the -product, we have (x1 x2 , eH ) = (x1 eH (x2 ), eH eH ) = (x1 , eH ) (x2 , eH ) . Note, here we use the fact that eH = IdN , which is again a consequence of being a group homomorphism. A similar argument works for H G . Lets now identify N and H with their images (N, eH ) and (eN , H ) in G. Then it is obvious that N H = {eG } in G, where eG := (eN , eH ) is the identity element in G. On the other hand, we have G = N H , because every element (x, h) G can be written as (x, h) = (x eN , eH h) = (x eH (eN ), eH h) = (x, eH ) (eN , h) , where (x, eH ) N and (eN , h) H . Thus, we conclude that G N = H.
Problem 3. First, we observe that N H = {e} in G. Indeed, N H is a subgroup in both N and H , and hence, by Lagranges Theorem, its order |N H | must
divide both |N | and |H | . But |N | and |H | are relatively prime. Hence, |N H | = 1 , meaning that N H = {e} . Now, take any elements x N and h H and consider their commutator [x, h] := x h x1 h1 in G. Since H is normal, we have h H x h x1 H and h1 H . Whence [x, h] = (x h x1 )h1 H . On the other hand, N is also normal, so x N x1 N h x1 h1 N , whence [x, h] = x (h x1 h1 ) N . Thus, we see that [x, h] N H and therefore [x, h] = e . Problem 4. It is straightforward to check that the compostion of two functions f1 (z ) = (a1 z + b1 )/(c1 z + d1 ) and f2 (z ) = (a2 z + b2 )(c2 z + d2 ) is given by (f1 f2 )(z ) := f1 (f2 (z )) = So, letting ab cd f (z ) = az + b , cz + d (a1 a2 + b1 c2 )z + (a1 b2 + b1 d2 ) . (c1 a2 + d1 c2 )z + (c1 b2 + d1 d2 )
and taking into account that the assumption ad bc = 0 we impose on f (z ) ab is equivalent to the matrix being invertible, we get a surjective cd group homomorphism GL(2, C) F (C) . Its kernel consists of matrices which correspond to the identity function f (z ) = z . These are exactly the 0 scalar matrices with = 0. By the First Isomorphism Theorem, 0 we conclude that F (C) GL(2, C)/C , where C is identied with a sub= group of (nonzero) scalar matrices in GL(2, C) . Problem 5. This is an immediate consequence of Theorem 2.133 on p. 193. Problem 6. Since |H | = p is prime, H must be a cyclic group (see Corollary 2.87 on p. 157). Now, due to HW Problem 2.94 on p. 171 (see also your lecture notes), we know that the order of the automorphism group Aut(H ) of H is |Aut(H )| = (p) = p 1 .
On the other hand, since H is normal in G, we can dene a map f : G Aut(H ) by g Adg , where Adg (h) = g h g 1 for h H . As shown in Problem 2(b) above, this map is a group homomorphism. Since G is a p-group, the image of f is also a p-group, so that |Im(f )| = pk for some k 0. By Lagranges Theorem, |Im(f )| must divide |Aut(H )| = p 1, which is possible only if k = 0. Thus, Im(f ) is a trivial subgroup of Aut(H ) (consisting only of the identity map IdH ). In other words, we have Adg = IdH for all g G. This means that Adg (h) = h, and hence g h = h g for all g G and all h H . Whence H Z (G) .