# Solutions of exercise sheet 4

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1 D-MATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 4 The content of the marked exercises (*) should be known for the exam. 1. Prove the following two properties of groups: 1. Every subgroup of a cyclic group is cyclic [Recall, we say that a group G is cyclic if G = g for some g G. A cyclic group can be either finite or infinite.] 2. Given a group G, if Aut(G) is cyclic then G is abelian [Hint: Consider the conjugation map G Aut(G).] 1. Suppose G is a cyclic group and H G. Let G = g. If H = {1 G }, then it is cyclic (generated by 1 G ) and we are done. Else there exists m Z \ {0} such that x = g m H. Up to inverting x, we may assume that m > 0. Then we can assume without loss of generality that m is the minimal positive integer such that g m H. We claim that H = g m, which makes H cyclic. The inclusion follows from the facts that g m H and H G. For the inclusion, pick an element g s H. Then, diving s by m in Z, i.e., finding k Z, 0 r < m such that s = km + r, we obtain g s = (g m ) k g r, that is g r = g s (g m ) k H. By minimality of m, we obtain r = 0, so that g s = (g m ) k g m and H g m. 2. Now suppose G is a group with Aut(G) cyclic. Then, considering the conjugation morphism γ : G Aut(G) sending g to the inner automorphism x gxg 1, we have that ker(γ) = Z(G) (see previous Exercise sheet s solution, Exercise 6), so that G/Z(G) = Im(γ) Aut(G), and by previous point G 1 := G/Z(G) is cyclic. Take a generator tz(g) for G 1, by fixing a suitable t G. Then every element x G can be written as x = t n c, with c Z(G) and n Z. We now prove that t Z(G), so that Z(G) = G and G is abelian. Let x G and take the commutator [x, t] = xtx 1 t 1. Writing down x = t n c, we get [x, t] = t n ctc 1 t n t 1 = t n+1 cc 1 t n 1 = 1 G because c Z(G). Hence t Z(G) = G and we are done. 2. Let H, K be subgroups of G, and assume that hk = Kh for every h H. 1. Show that: H K H; Please turn over!

2 HK G; K HK. 2. Prove that there is an isomorphism H/(H K) HK/K [Hint: Define first a group homomorphism H HK/K] 1. Assume x H K and h H. By hypothesis, hx = yh for some y K, and being y = hxh 1 H, we obtain hxh 1 H K, proving H K H. Now we prove that HK G using subgroups criterium: 1 G = 1 G 1 G HK; Given elements h i H and k i K, i = 1, 2 one has h 1 k 1 (h 2 k 2 ) 1 = h 1 k 1 k 1 Then k 1 k2 1 h 1 2 Kh 1 2 = h 1 2 K, so that we can write k 1k2 1 h 1 2 = h 1 2 k 0, with k 0 K, implying h 1 k 1 (h 2 k 2 ) 1 = h 1 h 1 2 k 0 HK. K is clearly contained in HK (as the set of elements of the form 1 G k, with k K), hence K HK. Moreover, k K and x HK, x = h 0 k 0, we have xkx 1 = h 0 k 0 kk0 1 h 1 0 hkh 1 = K by hypothesis, and we conclude that K HK. 2 h We define the group map f : H HK/K sending h hk. It is easily seen to be surjective: every element in HK/K has the form hkk with h H and k K. But hkk = hk = f(h). Then applying the First Isomorphism Theorem we get an induced isomorphism f : H/ ker(f) HK/K defined by f(h ker(f)) = hk, where ker(f) = {h H : hk = K} = H K, so that f is the required isomorphism. 3. Let G be a group with a normal subgroup H G and consider the canonical projection π : G G/H sending g gh. Prove the following statements: 1. If K G/H, then π 1 (K) is a subgroup of G containing H. 2. Conversely, if we have an intermediate subgroup H K G, then π(k ) G/H. 3. The map is a bijection. f : { subgroups K, H K G } K π(k ) { subgroups K G/H 4. For every H K G, one has that K G if and only if f(k ) G/H. 1. Since π is a group homomorphism, we have that π 1 (K) G by Exercise 1.4 from Exercise Sheet 2 (where the answer was yes ). Moreover, for h H we have that π(h) = 1 G/H K, so that h π 1 (K), and H π 1 (K). } See next page!

3 2. We have that π(k ) = Im(π i K ), where i K : K G is the canonical inclusion of K in G (a group homomorphism). Hence π(k ) G/H. 3. Let us prove that the map g in the other direction defined via g(k) = π 1 (K) is an inverse of f: f g = id: Let K G/H. Then (f g)(k) = π(π 1 (K)) = {π(x) x G, π(x) K} = K Im(π) = K, being π surjective. Hence f g = id. g f = id: Let H K G. Then (g f)(k ) = π 1 (π(k )) = {x G : π(x) = π(y), y K } = {x G : xh = yh, y K } = {x yh, y K } = K H = K, being H K. Hence f g = id. 4. Suppose that K G. Then we have a unique group homomorphism p K,H : G/H G/K such that p K,H π = π K, where π K is the canonical projection π K : G G/K. This homomorphism p K,H is the one sending gh gk, which is well-defined because H K = ker(π K : G G/K ), so that the ambiguity of taking gh G/H, which lies in H does not change the image of the element. Then f(k ) = π(k ) = {k H : k K } = ker(pi K,H) G/H. Conversely, suppose that f(k ) = π(k ) G/H. Then by previous point we have K = π 1 (π(k )) = π 1 (f(k )) G because the counterimage of a normal subgroup via a group homomorphism is always normal subgroup, and we are done [The proof that given a group homomorphism α : G 1 G 2 and K 2 G 2 we get α 1 (K 2 ) G 1 can be done as follows: consider the canonical projection π 2 : G 2 G 2 /K 2. Then f 1 (H 2 ) = {x G 1 f(x) H 2 } = {x G 1 f(x) ker(π 2 )} = = {x G 1 (π 2 f)(x) = 1 G2 /K 2 } = ker(π 2 f) G 1.] 4. Let G be a group and H G with [G : H] = 2. Prove: H G. Being [G : H] = 2, we have H G, and we can take g G \ H. Then H and gh are two disjoint right cosets of H, while H and Hg are two disjoint left cosets of H. Since [G : H] = 2, we have H gh = G = H Hg, and disjointness gives gh = G\H = Hg. So now let x G. If x H, i.e. xh = H, i.e. H = Hx, then xh = Hx. Else, xh = gh = Hg = Hx. Hence xh = Hx in any case, and H G. 5. (*) Let A be a simple finite abelian group. 1. Show that A is generated by an element x A different from 1 A. 2. Show that A = Z/kZ where k is a prime. Conversely, show that Z/pZ is a simple group for every prime number p. Please turn over!

4 1. Suppose M A. Then being A abelian, for each a A one has aha 1 = {aha 1 h H} = H, so that H A. This means that, being A simple, M = A or M = {1 A }. Also, by definition of simple group we have A {1 A }, and we can take x A, x 1 A. Look at M = x. Being 1 A x M, we get M {1 A }, so that M = A. Hence A is generated by (any element) x 1 A. 2. Let x A, x 1 A. Then we have just proved that the map ϑ : (Z, +) A sending n a n is surjective. It is easily seen to be a group homomorphism so that A = Z/ ker(ϑ). By Exercise 1.1, we have that ker(ϑ) is a cyclic subgroup of (Z, +), implying ker(ϑ) = kz, where we can k can be taken positive. We can exclude the cases k = 0 (which gives A = [Z : 0Z] =, contradiction with A finite) and k = 1 (which gives A = {1 A }). As, seen in the last point, every non-trivial element generates A, so that all the non-trivial elements have the same order k. This implies that k is prime. Else, there would be some proper divisor 1 d k of k, and for x A of order k we would have that the order of x d A \ {1 A } is k/d. Hence A = Z/kZ, with k a prime number. Conversely, every group Z/pZ, with p prime number is simple, since it has p > 1 elements, and by Lagrange Theorem every subgroup needs to have order dividing p, that is, order 1 or p. 6. (*) Given two group homomorphisms α : H G and β : G K we say that H α G β K is an exact sequence if Im(α) = ker(β). Moreover, given group morphisms ( ) α n 2 α n 1 α G n 2 G n 1 G n α n+1 n Gn+1 G n+2 α i 1 α we say that ( ) is an exact sequence if G i 1 G i i Gi+1 is an exact sequence for every i. We denote by 1 the trivial group {1}. Notice that for every group G there exists a unique homomorphism 1 G and a unique homomorphism G Prove that for any group homomorphism f : G H one has: 1 G f H is an exact sequence if and only if f is injective; G f H 1 is an exact sequence if and only if f is surjective. 2. We call a short exact sequence any exact sequence of groups of the form 1 H G K 1. Show that given the exact sequence above, there exists a subgroup H G such that H = H and K = G/H. See next page!

5 1. We have that 1 G f H is an exact sequence if and only if ker(f) = Im(1 G) = 0, if and only if f is injective. Moreover, G f H 1 is an exact sequence if and only if Im(f) = ker(h 1) = H, if and only if f is surjective. 2. Let α : H G and β : G K be the given maps. Call H := α(h) via the first map. Being α injective by previous point, we have that the map H H sending h α(h) is an isomorphism H = H. Using exactness at G, we have H = ker(β). Then applying First Isomorphism Theorem to the map β, which is surjective by previous point, we obtain K = G/ ker(β) = G/H and we are done.

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