Practice Test for Chapter 7. I. Specific Groups. (1) Show that any Quotient Group of Z is finite cyclic. First show that any subgroup of Z is cyclic. Let H be some subgroup of Z. Let n be the least positive integer in H. Let m be any other integer in H. Then by division algorithm, m = nq + r where r is either 0 or positive and < n. Since n is in H and m is in H, then m - nq = r is in H. -1 < r < n. But n is the least positive integer in H. Therefore r = 0, and m is a multiple of n. That is, every element in H is a multiple of n, the least positive integer in H. H = . Every quotient group of Z is Z/n for some integer n. The cosets in Z/n are , 1+, 2+,...(n-1)+. 1+ has order n in the quotient group Z/n. Z/n has n elements. The coset 1+ generates all the cosets of Z/n. Z/n is cyclic. (2) Show that any finite group is isomorphic to a subgroup of S_n. Let G be a finite group of n elements. Let H be the subset of S_n defined as follows: H consists of permutations of the elements of G. Define p_g in H where g is in G by the following. p_g (b) = b*g for all b in G, where * is the group multiplication of G. Consider the mapping theta: G -> H defined as follows. theta(g) = p_ [ theta(g) ] (b) = b * g. To show that theta is one to one, theta(g_1) = theta(g_2) implies p_g1 = p_g2 implies for all b in G that b * g1 = b * g2 implies g1 = g2. To show that theta is onto, Let p_b be an arbitrary element of H. By definition of theta, theta (b) = p_b. To show that theta preserves the group multiplication, Theta( g_1 * g_2) = p_(g_1*g_2) Theta( g_1 * g_2) (b) = b *( g_1 * g_2) for all b in G. = ( b * g_1) * g_2 for all b in G. =[ theta(g_1) (b) ] * g_2 for all b in G = ( [theta)g_2 ] ( [ theta(g_1) (b) ]) = [ theta(g_1) * theta(g_2) ](b) for all b in G. Theta (g_1 * g_2) = theta(g_1) * theta(g_2) (3) List all the elements of A_4 as a product of disjoint cycles, and find all the normal subgroups of A_4. A_4 = {(1), (12)(34), (13)(24), (14)(23), (123), (124), (132), (134), (142), (143), (234), (243)} The only proper normal subgrop of A_4 is {(1), (12)(34), (13)(24), (14)(23)}. To see this, we let H be a normal subgroup of A_4. We show that if H contains one 3 cycle, then it contains all 3 cycles of A_4. Suppose H contains the 3 cycle (abc) which represents any 3 cycle in A_4. Let d represent the element of {1,2,3,4} not in the cycle (abc). Since H is normal it also contains (ad)(abc)(ad) = (a)(bcd)=(bcd) (bd)(abc)(bd) = (adc) (cd)(abc)(cd) = (abd) (abc)^-1 = (acb) (bcd)^-1 = (bdc) (adc)^-1 = (acd) (abd)^-1 = (adb) This has shown that if H is a normal subgroup that contains one 3 cycle it contains all 3 cycles. The 3 cycles generate all of A_4. H = A_4. Suppose H contains one of {(12)(34),(13)(24),(14)(23)} Represent the element in H of this form by (ab)(cd). Then since H is normal, H also contains (ac) [ (ab)(cd) ] (ac) = (ad)(bc) (ad) [ (ab)(cd) ] (ad) = (ac)(bd) Thus if H contains any of { (12)(34), (13)(24), (14)(23) }, it contains all of them. The only proper normal subgroup of A_4 is H = {(1), (12)(34), (13)(24), (14)(23)}. The normal subgroups of A_4 are {(1)}, {(1), (12)(34), (13)(24), (14)(23)}, A_4. (4) Prove that no subgroup of order 2 in S_n (n > 2) is normal. Let H be a subgroup in S_n of order 2. Then H = { (a),(ab) } where a and b are distinct elements from {1,2,...n}. Let c be an element from {1,2,3...n} that is different from a and b. (ac)(ab)(ac) = (a)(bc) = (bc) is not in H. Therefore H is not normal. No subgroup of order 2 can be normal in S_n if n > 2. (5) Show that up to conjugation, there is only one k-cycle in S_n for every k = 1,2,3....n. Let (a1a2a3...ak) and (b1b2b3...bk) represent two arbitrary k cycles in S_n. Then [(a1b1)(a2b2)(a3b3)...(akbk)] (a1a2a3...ak) [(akbk)...(a3b3)(a2b2)(a1b1)] is the conjugation of (a1a2a3...ak) that is equal to (a1)(a2)(a3)...(ak)(b1b2b3...bk) = (b1b2b3...bk). Thus any two k-cycles in S_n are conjugates of one another. Up to conjugation there is only one k-cycle in S_n, for any k = 1,2,3...n. II. Subgroups. (1) If H is any subgroup of G, not necessarily normal, then there is a bijection between the set of left cosets and the set of right cosets. Also given the element a in G, there is a bijection between aH and Ha. Define f to map the left cosets into right cosets by f( aH) = H a^-1. To show that f is well defined and that f is one to one, aH = bH if and only if b^-1 a H = H if and only if b^-1 a is in H if and only if a^-1 b is in H if and only if H a^-1 b = H if and only if H a^-1 = H b^-1 if and only if f(aH) = f(bH) To show that f is onto, let Hb be a right coset. Then the left coset [b^-1] H is mapped onto Hb by f. f is onto. f is a well defined bijection of left cosets onto right cosets. To show that, given the element a in G, there is a bijection between aH and Ha, proceed as follows: Define r: H -> Ha by r(h) = ha for each h in H. Define l: H -> aH by l(h) = ah for each h in H. Both r and l are bijections since r(h1) = r(h2) implies h1 a = h2 a implies h1 = h2. r is 1-1. l(h1) = l(h2) implies a h1 = a h2 implies h1 = h2. l is 1-1. Given k_r in Ha implies there exist h_r in H such that k_r = h_r a implies k_r = r(h_r). r is onto. Given k_l in aH implies there exist h_l in H such that k_l = a h_l implies k_l = l(h_l). l is onto. Both r and l are bijections. Then r(l^-1) is a bijection from aH to Ha. (2) Show that if H is a finite nonempty subset of the group G which is closed under the group operation, then H is a subgroup. Let b be some non-identity element in H. Consider the set of consecutive powers of b = { b, b^2, b^3, ...} Since H is finite, there must be some repetition of values, say for i< j that b^i = b^j. Then if c is any element in H, c b^j = c b^i c b^(j-i) = c b^(j-i) is the identity in H. also since [ b ] [ b^(j-i-1] = b^(j-i) is the identity and [ b^(j-i-1)] b = b^(j-i) is the identity, b^(j-i-1) is the inverse of b. H is a subgroup. (3) If the element a in G has order n, what is the order of a^t? Give a formula for order(a^t) in terms of t and n. Justify your answer. n = order(a) n is the least non-negative integer such that a^n = identity element. Let k = order(a^t). k is the least non-negative integer such that (a^t)^k = identity element. k is the least non-negative integer such that a^( t k) = identity element. k is the least non-negative integer such that t k is a multiple of n. t k =t n/gcd(t,n) k = n/gcd(t,n) order(a^t) = order(a) / gcd( t, order(a) ) (4) if ab is in the center of group G, then prove that ba is also in the center of group G. ab commutes with every element of G. ab commutes with a^-1. [ ab] [a^-1] = [a^-1] [ab] = b ab [ a^-1] a = ba ab = ba III. Miscellaneous (1) Let G be any group. From the group axioms, show that the identity element is unique, and that if a is in G that a^-1 is unique. Suppose e and f are both identity elements. Then e f = e since f is an identity element. and e f = f since e is an identity element. Therefore, e = e f = f. The identity element is unique. Suppose b and c are both inverses of the element a. Then ba = identity and ac = identity. and b = b ( ac ) = (ba) c = c. The inverse of a is unique. (2) If a and b commute, and order(a) = p, and order(b) = q, with p and q not necessarily distinct or prime, what is the order(ab)? Let n = order(ab) n is the least non-negative integer such that (ab)^n = identity. a and b commute. Let e be the identity element. (ab) ^ [ pq/gcd(p,q) ] = a^[pq/gcd(p,q)] b^[pq/gcd(p,q)] = [ a^p ]^(q/gcd(p,q)) [b^q]^(p/gcd(p,q)) = e^(q/gcd(p,q) ) e^(p/gcd(p,q) ) = e * e = e n divides pq/gcd(p,q). n = order ( ) (ab)^p = a^p b^p = b^p (ab)^p is an element of order( (ab)^p ) = order( (b^p) ) = p/gcd(p,q) p/gcd(p,q) divides n (ab)^q is an element of order( (ab)^q ) = order( a^q b^q ) = order( a^q ) = q/gcd(p,q) q/gcd(p,q) divides n lcm( p/gcd(p,q) , q/gcd(p,q) ) divides n pq/[ gcd(p,q) ]^2 divides n. n divides pq/gcd(p,q). To show that no better result is possible, look at the example of z mod 210, with p = 30 and q = 105. a b a+b n 49 4 53 210 7 2 9 70 77 8 85 42 7 8 15 14 (3) Produce a list of Groups such that every homomorphic image of Z_8 is homomorphic to exactly one group on the list. z_8 is isomorphic to Z_2 X Z_2 X Z_2 list = group of single element Z_2 Z_2 X Z_2 Z_2 X Z_2 X Z_2 (4) Give an example of a group G and a normal subgroup N in G such that N and G/N are abelian, but G is not an abelian group. Let G = S_3 = { (1),(12),(13),(23),(123),(132)} G is not abelian. Let N = {(1),(123),(132)} N is an abelian group of order 3. G/N = { { (1),(123),(132) }, { (12),(13),(23) } } is an abelian group of order 2.