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(math311)[2008](f)quiz(v1)~65^_10502.pdf
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Solutions for Quiz (version 1) for Math 311, 2008

9
Problem 1 (30 points) Compute the order of elements. (1). e i in C. .
(2). iin C. . (3). 4 in Z14. (4). (123)(679) in S9. (5). . 0 1 .10 . in
GL(2,R).

Solution: (1) 18. (2) 4. (3) 7. (4) 3 . (5). 4.
. 123456 .
Problem 2 (25 points) Let = S6 be an odd 235 xy 4
permutation.
(1). Compute xand y. (2). Compute .1 .
(3). Decompose into a product of disjoint cycles.
(4). Compute the order of .
(5). Compute 12 .
Solution: (1) x=1,y =6. . 123456 .
(2)
.1 = 412635

(3)
= (123564).

(4)
The order of is 6.

(5)
12 = e.


2
Problem 3 (25 points) Mark each of the following true or false (just answer
true or false, no reasons needed).
(1). If S8 is a transposition, then has order 2.
(2). If S8 is has order 2, then is a transposition.
(3). If H is a subgroup of a .nite group G, then |H|is a divisor of |G|.
(4). If H is a subgroup of a .nite group G, then for all a,b G, |aH|= |bH|.
(5). If G is an ablian group, then G is a cyclic group.

Solution: (1) true. (2) false. (3) true. (4) true. (5) .ase.
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). If |G|.= 1, then |G|is 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.