Download Homework 9 - Material from Chapters 9-10

Math 400
SOLUTIONS
Homework 9 - Material from Chapters 9-10
1. What is the order of the factor group Z60 /h15i?
Solution: Since |Z60 | = 60 and |h15i| = 4, the order of the factor group is 60/4 = 15.
2. What is the order of the factor group (Z12 ⊕ U (12)/h(2, 7)i?
Solution: Note |U (12)| = φ(12) = 4. The order of the subgroup h(2, 7)i is the order
of the element (2, 7), which is the LCM of the order of 2 (in Z12 ) and the order of 7
(in U (12)), so the LCM of 6 and 2, which is 6. Thus the order of the factor group is
ds 12·4
= 8.
6
3. What is the order of the element 14 + h8i in the factor group Z24 /h8i?
Solution: This question is asking how many times the coset 14 + h8i must be added
to itself in order to get the identity coset 0 + h8i. To equal the identity coset we must
have a + h8i where a ∈ h8i, so a = 0, 8, or 16. We can just add 14 to itself in Z24 until
we get 0 8, or 16, and see how many times it takes. So: 14 + 14 = 4, 14 + 14 + 14 =
18, 14 + 14 + 14 + 14 = 8. Thus 4 times (14 + h8i will be 8 + h8i = 0 + h8i, so the order
of this element in the factor group is 4.
4. Prove that any factor group of a cyclic group is always cyclic. (But note that a cyclic
factor group does NOT guarantee a cyclic “original” group.)
Solution: Suppose G is a cyclic group, so G = hxi for some x ∈ G. Let N be any
normal subgroup of G. We must show that G/N is a cyclic group. Take an arbitrary
element of the factor group G/N , let’s say aN . Since a ∈ G, a = xk for some k ∈ Z.
So aN = xk N = (xN )k . Since aN was an arbitrary element of G/N , every element of
G/N is some power of xN , so G/N = hxN i, so G/N is cyclic. 5. Prove that any factor group of an Abelian group is always Abelian. (But note that an
Abelian factor group does NOT guarantee an Abelian “original” group.)
Solution: Suppose G is an Abelian group, and N is any normal subgroup of G. We
must show that G/N is Abelian. Let aN, bN ∈ G/N be arbitrary elements of the
factor group. Then a, b ∈ G and since G is Abelian we know ab = ba. Thus we have
(aN )(bN ) = abN = baN = (bN )(aN ), so the cosets (elements of the factor group) aN
and bN commute. Since they were arbitrary elements of the factor group, all elements
of G/N commute, so G/N is Abelian. 6. Let G = R∗ and N = R+ (both under multiplication. Since G is Abelian, N C G.
Write out the elements of G/N , and a Cayley table for this factor group. Looking at
your Cayley table, can you think of a familiar group that G/N is isomorphic to?
Solution: The elements of G/N are the cosets of N in G. There are only 2 cosets: N
itself (which is all positive real numbers) and −1N (which is all negative real numbers).
With brief calculations we get the following Cayley table:
N
−1N
+ −
N
N
−1N , which we can also think of as + + −
−1N −1N
N
− − +
Math 400
SOLUTIONS
This is clearly the same thing as the Cayley table for the group {1, −1} under multiplication (or equivalently Z2 , with N playing the role of 0 and −1N playing the role
of 1). So this factor group R∗ /R+ is isomorphic to {1, −1} (or equivalently, Z2 ).
7. Let G be a group with center Z(G). Prove that it is impossible for |G : Z(G)| to be
prime. (Hint: use Theorem 9.3.)
Solution: Suppose for contradiction that |G : Z(G)| = p, some prime number. (So
p > 1.) Now we know Z(G) C G and therefore we can consider the factor group
G/Z(G). The order of this factor group equals the index, so the order is p. We know
(from chapter 4 stuff) that any group of prime order is cyclic, so G/Z(G) must be cyclic.
Then from theorem 9.3, we know G is Abelian. But if G is Abelian, all its elements
commute with all its elements, so Z(G) = G. Butg then |G : Z(G)| = |G : G| = 1,
contradicting that this index equals p > 1. Therefore |G : Z(G)| is never prime. 8. Suppose H ≤ G and H is the only subgroup of G that has order |H|. Prove that
H C G.
Solution: Let x ∈ G be arbitrary. Consider xHx−1 . (This is the set of all elements
of the form xhx−1 , where h ∈ H.) This set forms a subgroup of G: if xhx−1 , xkx−1 ∈
xHx−1 , their product is xhx−1 xkx−1 = xhkx−1 ∈ xHx−1 , since H is closed so hk ∈ H.
Thus xHx−1 is closed. Likewise, (xhx−1 )−1 = xh−1 x−1 by socks-shoes, so xHx−1
contains an inverse for each of its elements. Therefore xHx−1 ≤ G. Note that it has
the same order as H since the order of H = the cardinality of any of its cosets xH =
the cardinality of the set xHx−1 by the same reasoning as in chapter 7.
Since H is the only subgroup of G with its order, H = xHx−1 and therefore H C G.
9. For each function below, determine whether it is a homomorphism, and if so find its
kernel.
(a) For any group G, ϕ : G → G given by ϕ(x) = x−1
Solution: Not a homomorphism: ϕ(ab) = (ab)−1 = b−1 a−1 , which is not necesarily the same as a−1 b−1 = ϕ(a)ϕ(b). (On the other hand, if G is Abelian this is
a homomorphism – in fact an isomorphism.)
(b) ϕ : R∗ → {1, −1} under multiplication, defined by ϕ(x) = 1 if x is positive and
−1 if x is negative
Solution: Yes, it’s a homomorphism. Since we know positive times positive
equals negative times negative equals positive, and negative times positive equals
positive times negative equals negative (all for real numbers) it follows that as
defined this function will always have ϕ(a)ϕ(b) = ϕ(ab). The kernel is all elements
of R∗ that map to 1, which means it is all positive real numbers, R+ .
(c) For any Abelian group G and any integer n > 1, ϕ : G → G given by ϕ(x) = xn
Solution: Yes, it’s a homomorphism: If G is Abelian then ϕ(ab) = (ab)n =
an bn = ϕ(a)ϕ(b) for all a, b ∈ G. The kernel is {a ∈ G | an = e}, which is all the
elements of G whose orders divide n.