M417 Practice Exam 1 Spring 2004

Note: This is the Exam 1 I gave when I taught this class last year. We had gotten a little further; our Exam 1 next Monday won't cover anything beyond Chapter 4 (so you don't have to worry about a problem like # 6 below). Like last year's exam, this year's will have a mix of problems, some right from the homework, some similar to homework problems or things done in class, and some new ones, and you will get a choice. I will post answers later this week.

Instructions: Do any four of the eight problems. Don't forget to put your name on your answer sheets.
1. Let h: A -> B be a function from a set A to a set B. If h: A -> B is surjective, show that h-1 : 2B -> 2A is injective.

See the solutions to homework 5 for the answer to this one.

2. Let G be a group which has exactly three different subgroups, including a proper subgroup H of order 7. Show that G is cyclic, and determine |G|.

The subgroups of G are < e >, H and G. Let g be in G-H. Then < g > can't be < e > or H, so < g > = G, hence G is cyclic. Thus 7=|H| divides G. Say |G|=7m. If m=1, then H=G, which is false. If 1 < m but m is not 7, then G has a proper subgroup M of order m, hence M is not < e > nor H, contradicting there being only two proper subgroups. Thus m=7 and |g|=49.

3. Let {f1, f2, ... } be a sequence of integers such that f1 > 1 and for i > 0 such that fi+1 >= 2fi - 1. Prove that fn >= (2n + 2)/2 is true for all integers n >= 1.

Since f1 is an integer and f1 > 1, then certainly fn >= (2n + 2)/2 holds for n = 1. Now assume that i >= 1 and that fn >= (2n + 2)/2 holds for n = i. Then fi+1 >= 2fi - 1 >= 2(2i + 2)/2 - 1 = 2i + 1 = (2i+1 + 2)/2, hence by induction we conclude that fn >= (2n + 2)/2 is true for each integer n > 0.

4. Let g and h be elements of a group G. Note that G need not be finite. Prove that |ghg-1| = |h|. (Hint: show that (ghg-1)i = ghig-1.)

First, note that (ghg-1)i = (ghg-1)(ghg-1)...(ghg-1) = gh(g-1g)h(g-1g)...(g-1g)hg-1 = ghh...hg-1 = ghig-1. If for some positive integer i we have (ghg-1)i = ghig-1 = e, then hi = g-1ghig-1g = g-1eg = e, so h has finite order. And if h has finite order, then for some positive integer i we have hi = e, then we have (ghg-1)i = ghig-1 = geg-1 = e. Thus h has finite order if and only if ghg-1 does (i.e., if h has infinite order, then so does ghg-1, and vice versa).

Now say h has finite order i. Then as before (ghg-1)i = ghig-1 = e, then hi = g-1ghig-1g = g-1eg = e, so | ghg-1 | <= i = | h |. Note that h = fxf-1, where x = ghg-1 and f = g-1. But we just saw that | fxf-1 | <= | x |, hence | h | = | fxf-1 | <= | x | = | ghg-1 |. Thus | h | = | ghg-1 |.

5. Let G be an abelian group. Let n > 0 be an integer. Let N be the subset of G of all elements g of G such that gn = e. Prove that N is a subgroup of G.

Since e is in N, N is nonempty. If g and h are in N, then (gh)n = gnhn, since G is abelian, so (gh)n = ee = e, hence gh is in N and so N is closed under the group operation. And if g is in N, then e-1 = (gn)-1 = (g-1)n, hence g-1 is in N, so N is closed under taking inverses. Thus N is a subgroup.

• Write (1 2 3 4 5 6)3 as a product of disjoint cycles.
• Write (1 2 3)(1 2)(3 4) as a product of disjoint cycles.
• Find |(1 2 3)(1 2)|; show how you find your answer.

We'll skip this one.

6. Consider the group Z899 of integers modulo 899. Note that 899 = 29x31. For each positive integer n, determine the number of elements of Z899 of order n. Explain how you obtain your answer.

This applies Corollary 1 of Theorem 4.1 and Theorem 4.3 (which implies that orders of elements of cyclic groups divide the order of the group) and Theorem 4.4 (which implies that for each divisor d of |Z899| = 899, there are phi(d) elements of of order d). Thus there are no elements of order n unless n is 1, 29, 31 or 899. There is phi(1) = 1 element of order 1, there are phi(29) = 28 elements of order 29, phi(31) = 30 of order 31 and phi(899) = phi(29)xphi(31) = 28x30 = 840 of order 899.

7. Give an example of a group G and a nonempty subset H of G such that whenever a is in H and b is in H, then ab is in H, but nonetheless H is not a subgroup of G.

Take the positive integers for H and the integers for G. Since H does not include the inverses of its elements, it is not a subgroup of G.