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} : 2^{B} > 2^{A} 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 GH. 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 {f_{1}, f_{2}, ... }
be a sequence of integers such that
f_{1} > 1 and for i > 0 such that
f_{i+1} >= 2f_{i}  1.
Prove that f_{n} >= (2^{n} + 2)/2
is true for all integers n >= 1.
Since f_{1} is an integer and f_{1} > 1, then certainly
f_{n} >= (2^{n} + 2)/2
holds for n = 1. Now assume that i >= 1 and that
f_{n} >= (2^{n} + 2)/2 holds for n = i. Then
f_{i+1} >= 2f_{i}  1 >= 2(2^{i} + 2)/2  1
= 2^{i} + 1 = (2^{i+1} + 2)/2, hence
by induction we conclude that f_{n} >= (2^{n} + 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}
= gh^{i}g^{1}.)
First, note that (ghg^{1})^{i}
= (ghg^{1})(ghg^{1})...(ghg^{1})
= gh(g^{1}g)h(g^{1}g)...(g^{1}g)hg^{1}
= ghh...hg^{1} = gh^{i}g^{1}.
If for some positive integer i we have
(ghg^{1})^{i} = gh^{i}g^{1} = e, then
h^{i} = g^{1}gh^{i}g^{1}g
= g^{1}eg = e, so h has finite order. And if h has finite order,
then for some positive integer i we have
h^{i} = e, then we have
(ghg^{1})^{i} = gh^{i}g^{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} = gh^{i}g^{1} = e, then
h^{i} = g^{1}gh^{i}g^{1}g
= g^{1}eg = 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
g^{n} = 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} = g^{n}h^{n}, 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} = (g^{n})^{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 Z_{899}
of integers modulo 899. Note that
899 = 29x31. For each positive integer n, determine the number of
elements of Z_{899} 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
Z_{899}
= 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.