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.
- 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.
- 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.
- 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.
- 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 |.
- 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.
- 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.
- 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.