( ) 3 = ab 3 a!1. ( ) 3 = aba!1 a ( ) = 4 " 5 3 " 4 = ( )! 2 3 ( ) =! 5 4. Math 546 Problem Set 15

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1 Math 546 Problem Set 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 Suppose that a and b are elements of G show that aba!1 Solution: aba!1 ( ) 3 = aba!1 a =!e ba!1 a =!e ba!1 = abbba!1 = ab 3 a!1. ( ) 3 = ab 3 a!1. 3. (a). What is the order of U(5000)? Solution: U(5000) =!(5000) =! ( ) =! 5 4 ( )! 2 3 ( ) = 4 " 5 3 " 4 = (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 is divided by 19? Solution: By Fermat s Theorem, 2 18! 1mod19. Hence, 2 90! 1mod19. Thus, 2 100! 2 10 mod19. But simple division show that 2 10! 17mod19. Hence the remainder is (a). How many generators are there for Z 30? Solution: There are!(30) = 8 generators. (b). How many generators are there for (Z,!+)?

2 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. (b). What is the order of the element Solution: The order is 4. " 0!1 $ 1 0 ' (GL(2, R). & )! a b$, (c). Let H = * " c d & : ad ' bc = 1,!a,!b,!c,!d (R-. Show that H is a subgroup +. 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! a"g C(a)? Solution:! a"g C(a) = Z(G), the center of G.! (c). What is the Centralizer of 1 1 $ " 1 0 & 'GL(2, R)? )" a b, Solution: * $ b a! b ' : a,!b (R- Note this is a slightly different description + &. 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 )

3 Solution: Let x!c(a 3 ). Then x = xa 5 = xa 3 a 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. * - 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$! b b$ & " b b & =! 2ab 2ab $ " 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. 1 1! 2 2 $ It is simple to check that E = 1 1 & is an identity element for K. And for " 2 2 any A = a a 1 1! $! 4a 4a $ & 'K, the matrix A' = 1 1 & serves as an inverse for 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 = g 12 = e and so x has order 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 = b 11. 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 b 11. Thus a 6 = x = b 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

4 1 15. What is the order of the element 2 1+ i under multiplication? Solution: Note that x 2 = i,!x 4 =!1,!x 8 = 1. The order is 8. [ ] in the non-zero complex numbers 16. If a, b and c are elements of the group G, then what is the inverse of abc? Solution: abc ( )!1 = c!1 b!1 a!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]. Solution: [G : K] = G K,![G : H ] = G H,![H : K] = H K. Now just substitute. 19. What is the index of the subgroup {1, 1} in the Quaternions Q 8? Solution: 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.

5 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. Then e = g n = g mq+r = g m ( ) q g r! g r = ( g m ) "q H. Hence by the choice of m, r cannot be both positive and less than m. So r = 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.