=========================preview======================
(math311)[2008](f)quiz(v2)~65^_10503.pdf
Back to MATH311 Login to download
======================================================
Solutions to Quiz (version 2) for Math 311, 2008
7
Problem 1 (30 points) Compute the order of elements. (1). e i in C. .
.01.
(2). .i in C. . (3). 4 in Z10. (4). (12)(3479) in S9. (5). in 10
GL(2,R).
Solution: (1) 14. (2) 4. (3) 5. (4) 4. (5) 2.
. 123456 .
Problem 2 (25 points) Let = S6 be an odd 253 xy 4
permutation.
(1). Compute x and y. (2). Compute .1 .
(3). Decompose as a product of transpositions.
(4). Compute the order of .
(5). Compute 2 .
Solution: (1). x =6, y =1. . 123456 .
(2). .1 = . 513624
(3). = (125)(46).
(4). the order of is 6.
(5) 2 = (152).
2
Problem 3 (25 points) Mark each of the following true or false (just answer
true or false, no reasons needed).
(1). If S8 has order 3, then is a cycle of length 3.
(2). If S8 is a cycle of length 3, then has order 3.
(3). If H is a subgroup of a .nite group G, then |H|is a divisor of |G|.
(4). If H is a subset of a .nite subgroup G, and |H|is a divisor of |G|, then
H is a subgroup of G.
(5). If G is a cyclic group, then G is an ablian group.

Solution: (1) false. (2) true . (3) true. (4) false. (5) true.
Problem 4 (10 points) Let G be a .nite group. If every a G satis.es
a2 = e, prove that
(1). G is an abelian group.
(2). |G|is either 1 or an even number.

Proof. (1). For a,b G, ab G,so(ab)2 = e, i.e., abab = e, a(abab)b= ab.
Since a2 = e, b2 = e,wehave ba = ab. This proves G is abelian.
(2). Since |G|.= 1, there is a G, a .= e. Since a2 = 2, so the order of a is

2. By Theorem 10.12, 2 is a divisor of |G|. This proves |G|is even.
Problem 5 (10 points) Let H be a .nite subgroup of C. . Prove that H is Un for some n.
Proof. Let |H|= n. By Theorem 10.12, for every a H, an = 1. So H .Un. Since |H|= n = |Un|, we must have H = Un.