Hints for Homework 4
Chapter 2:
14
This is the problem I reassigned. Here's a mistake that some
people made last time: you can't choose just any a, b, and c and have
that ab = ca. The group has the property that for any a, b, c such
that ab = ca then b = c. You want to prove that the group is
abelian. Your goal is to show that xy = yx for any group elements
x, y. Match this equation up to b and c in the special property
and find an element a that makes the property applicable.
17
Gallian suggests to use Exercise #16. That seems like a good way to
start this. In that exercise, we have that (ab)^-1 = b^-1 a^-1, so
apply that to the statement in #17 and go from there.
36
Look at theorem 4.4 and its corollary on page 80 and try to apply
them.
Chapter 3:
11
When Gallian says find "a" group such that certain properties hold,
he doesn't mean the same group in each case. This wording may be
confusing, but his meaning is clear from context. The group element
ab has a well-defined order, which is a positive integer. The order
can't simultaneously be 3, 4, and 5. Note also that the order of
element ab doesn't have to be the same number for all choices of
a and b. For each case, find a group and two particular elements
of that group with order 2 and the property that |ab| = 3, 4, or
5, respectively.
14
Use the two-step subgroup test to show that the intersection is
closed under composition and inversion. That is, show that for
any two elements a, b that are in both H and K, that ab is in
the intersection and that a^-1 is in the intersection.
(Give a clear written explanation for this proof. For the
question "Can you see...", just give a brief answer.
Please think about it, but "no" is a valid response.)
17 and 18
Let the elements of the group be indexed as { a_1 = e, a_2,...,a_n }.
The row i column j entry (i,j) of the Cayley matrix is the element
a_i a_j . Given two elements a_i, a_j, if a_j is in the centralizer
of a_i, that is, in C(a_i), then a_i a_j = a_j a_i. This means
that the (i,j) entry is equal to the (j,i) entry.
24
Write out the conditions for an element to be in C(a) and C(a^3).
Show that they are equal by either (1) writing a reversible argument
that shows x is in C(a) iff x is in C(a^3), or (2) show that C(a) is
contained in C(a^3) and vice versa. One direction is easy. The other
direction is harder. Start with x a^3 = a^3 x and multiply on both
the left and right by a power of a that lets you cancel out a^5 and
show that x commutes with a different power of a. Then go from there.
(There are probably a couple of ways to do this problem.)
Chapter 4:
7 (contributed by Karen Reed)
For #7 you should answer two questions.
1. Why is the group not cyclic?
Look at U(8). Determine the orders of the elements.
Does U(8) = for any a in U(8)?
2. Look at the subgroups of U(8). Hint: Any two elements of
{3, 5, 7} will generate U(8). Ex. <3,5> 3*5 =15
3*5 = 7 mod8
16
Consider the factorization of 120 into primes as 2^3 * 3 * 5.
Apply the Fundamental Theorem of Cyclic Groups. The orders
of the groups in the chain must have must successively
divide each other (a_{i+1} divides a_i). If the quotient
of one order by the next has more than one prime factor, then
you're missing an intermediate step.
39 (contributed by Michelle Yakaboski)
Based on theorem 4.3 the fundamental theorem of cyclic groups, the
group, G, has as many subgroups as the number of divisors of the order
of the group. The number 32 has 6 divisors, that will work! 32 has the
divisors, 1, 2, 4, 8, 16, 32.
Therefore:
|a|=32, < a > = G
|a^2|=16, < a^2 > = {1, a^2, a^3...a^30}
|a^4|=8, < a^4 > = {1, a^4, a^8...a^28}
|a^8|=4, < a^8 > = {1, a^8, a^16, a^24}
|a^16|=2, < a^16 > = {1, a^16}
< a^32 > = < e >
A group with exactly 6 subgroups is < a^32 >, with the subgroups:
< a >, < a^2 >, < a^4 >, < a^8 >, < a^16 >, < e >.