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Math 546 Problem Set 15 1. Let G be a finite group. (a). Suppose that H is a subgroup of G and o(H) = 4. Suppose that K is a subgroup of G and o(K) = 5. What is H ! K (and why)? Solution: H ! K = {e} since H ! K is a subgroup of each of H and K and so its order must divide each of 4 and 5. So, H ! K = 1 . (b). Suppose that H is a subgroup of G and o(H) = 12. Suppose that K is a subgroup of G and o(K) = 18. Also H ! K " 4 . What is the value of H ! K (and why)? Solution: Similar to part (a), H ! K = 6 since that is the only number that satisfies the given conditions. (i.e., 6 ≥ 4 and 6 divides both 12 and 18. ( 2. Suppose that a and b are elements of G show that aba !1 ( Solution: aba !1 ) 3 ) 3 = ab 3a !1 . = ab a !1a b a !1a ba !1 = abbba !1 = ab 3a !1 . =!e =!e 3. (a). What is the order of U(5000)? Solution: U(5000) = ! (5000) = ! 5 4 2 3 = ! 5 4 ! 2 3 = 4 " 5 3 " 4 = 2000 . ( ) ( ) ( ) (b). Is U(15) cyclic? (c). Is Z 8 cyclic? Solution: Yes. Every Z n is cyclic – and is generated by any k in Z n that is relatively prime to n. (d). Is Z 4 isomorphic to the Klein 4-group? Solution: No. Z 4 is cyclic but the Klein 4-group is not cyclic. 4. What is the remainder when 2100 is divided by 19? Solution: By Fermat’s Theorem, 218 ! 1mod19 . Hence, 2 90 ! 1mod19 . Thus, 2100 ! 210 mod19 . But simple division show that 210 ! 17 mod19 . Hence the remainder is 17. 5. (a). How many generators are there for Z 30 ? Solution: There are ! (30) = 8 generators. (b). How many generators are there for (Z,!+) ? Solution: Two – the numbers 1 and –1 generate Z. 6. Try various values of n in the Group Therapy program and make some conjectures as to when U(n) is cyclic. 7. (a). Show that GL(2, R) is not an Abelian group. " 0 !1% (b). What is the order of the element $ ' (GL(2, R) . #1 0 & Solution: The order is 4. )!a b $ , (c). Let H = * # : ad ' bc = 1,!a,!b,!c,!d (R - . Show that H is a subgroup & +" c d % . of GL(2, R) . H is called the Special Linear Group SL(2, R) . Hint: for any two n ! n matrices A and B, det(AB) = det(A)det(B) . 8. Suppose that H is a subgroup of the group G and the order of G is 30. You check all the non-identity elements of G and find that there are three nonidentity elements a, b, and c that you know are in H and three d, f, and g that you are not certain of. None of the other non-identity elements belong to H. How many of d, f, and g belong to H? Explain. Solution: By LaGrange’s Theorem, either one or two of d, f, and g belong to H. 9. Let G be a group and a an element of G. The Centralizer of a is the set C(a) = { x !G : ax = xa} . (a). Show that C(a) is a subgroup of G. (b). What is ! C(a) ? a"G Solution: ! C(a) = Z(G) , the center of G. a"G !1 1 $ (c). What is the Centralizer of # & 'GL(2, R) ? "1 0 % b % )"a , Solution: * $ : a,!b (R - Note this is a slightly different description ' +#b a ! b & . than the one we did in class. (d). Suppose that G is a group of order 20. Could the Centralizer of G contain exactly three elements? Explain. Solution: No, C(a) is a subgroup of G and so its order must divide 20. (e). Suppose that a is an element of G with o(a) = 5 , and x !C(a 3 ) . Show that x !C(a 2 ) Solution: Let x !C(a 3 ) . Then x = xa 5 = xa 3a 2 = a 3 xa 2 ! a 2 x = a 2 a 3 xa 2 = a 5 xa 2 = xa 2 . There are many other ways to argue this. 10. Is (Q,!+ ) a cyclic group? Solution: No. (!a a $ + 11. Show that K = ) # : a 'R* , is a group under matrix multiplication. & *"a a % You need not verify that matrix multiplication is associative. [But it is not a subgroup of GL(2, R) .] What is the identity of K? Solution: Under matrix multiplication, any two matrices of the form in K, have a product of the form required to belong to K. More precisely, ! a a $ !b b $ ! 2ab 2ab $ # a a & #b b & = # 2ab 2ab & for any real numbers a and b. Thus K is closed " %" % " % under matrix multiplication. We already know that matrix multiplication is associative, so there is no need to check that. ! 12 12 $ It is simple to check that E = # 1 1 & is an identity element for K. And for " 2 2% !a a $ ! 14a 14a$ any A = # , the matrix serves as an inverse for A. 'K A' = & #1 1 & "a a % " 4a 4a% 12. Suppose that the order of the element g in the group G is 12. What is the order of x = g 3 ? Verify your answer. Solution: g k = e ! k " 12 , we know that none of x,!x 2 ,!x 3 can be e. However, x 4 = g12 = e and so x has order 4. 13. Suppose that a and b are elements of a group G, and o(a) = 12,!o(b) = 22 . Suppose that H =!< a > ! < b >!" {e} . (a). What is the order of H? Solution: The order of H is 2. (b). Show that a 6 = b11 . Hint: A non-trivial cyclic group has exactly one element of order 2. Solution: Let x be the non-identity element of H. Then x has order 2. However, the only element of order 2 in < a > is a 6 and the only element of order 2 in < b > is b11 . Thus a 6 = x = b11 . 14. Suppose that x is an element of the group G and x 8 = e . If x 2 ! e and the order of x is not 8, then what is the order of x? Solution: 4 15. What is the order of the element 1 [1 + i ] in the non-zero complex numbers 2 under multiplication? Solution: Note that x 2 = i,!x 4 = !1,!x 8 = 1 . The order is 8. 16. If a, b and c are elements of the group G, then what is the inverse of abc? !1 Solution: ( abc ) = c !1b !1a !1 17. (a). Suppose that H is a subgroup of U(56) . Given that 9 is in H, show that 25 is also in H. Solution: In U(56) , 9 2 = 25 . (b). What is the order of 5 in U(17)? – check with the group therapy software. (c). What is the order of 7 in U(15)? - check with the group therapy software. 18. The index of a subgroup H of the finite group G is the number of left cosets of H in G and is denoted by [G : H ] . So, by LaGrange’s Theorem, [G : H ] = o(G) o(H ) . Suppose that G is a finite group and H is a subgroup of G and K is a subgroup of H. Show that [G : K ] = [G : H ] ![H : K ] . G G H Solution: [G : K ] = ,![G : H ] = ,![H : K ] = . Now just substitute. K H K 19. What is the index of the subgroup {1, –1} in the Quaternions Q8 ? Solution: 4. 20. Show that if every element of G has order 2, then G is Abelian. Solution: See your notes. 21. Show that for any prime p > 2, there are exactly two elements of U(p) such that satisfy x 2 = 1 (i.e., there is exactly one element of order 2). Hint: If p is a prime and p divides the product nm of integers n and m, then p divides n or p divides m. Solution: Note that x 2 = 1 in U(p) is equivalent to x 2 ! 1mod p " p!divides!x 2 # 1 = (x + 1)(x # 1) . This in turn is equivalent to p divides x ! 1 or x + 1 . So x ! 1mod p or x ! "1mod p . 22. Let H be a subgroup of the finite group G and g an element of G with o(g) = n . Let m be the smallest positive integer such that g m !H . Show that m divides n. Hint: You know that n = mq + r where q and r are integers and 0 ! r < m . Solution: Suppose that n = mq + r . ( ) q ( ) "q Then e = g n = g mq + r = g m g r ! g r = g m #H . Hence by the choice of m, r cannot be both positive and less than m. So r = 0. 23. Show that if x,!y !G and xy !Z(G) , then xy = yx . Note: Recall that Z(G) denotes the center of G. Hint: Note that xy = yx ! xyx "1 y "1 = e . Solution: See your notes. 24. No group can have exactly two elements of order 2. Hint: Consider the two cases where ab = ba and ab ≠ba. Solution: Suppose that a and b are elements of order 2 in G. Then if ab = ba then, ab is a third element of order 2, while otherwise, aba is a third element of order 2.