114 Groups and Rings 2008–09
Suggested solutions to exercise set 2
February 16, 2009
20.1 We are asked to write down θ([a]) ∈ Sym(Z5 ) for each [a] ∈ Z5 , where
θ([a])([b]) = λ[a] ([b]) = [a] ⊕ [b] = [a + b] for [a], [b] ∈ Z5 .
Since θ is a homomorphism
it must sent the
identity element to the identity
[0] [1] [2] [3] [4]
element, so θ([0]) =
is the identity permutation. Now
[0] [1] [2] [3] [4]
[0]
[1]
[2]
[3]
[4]
[0] [1] [2] [3] [4]
θ([1]) =
=
[0 + 1] [1 + 1] [2 + 1] [3 + 1] [4+ 1]
[1] [2] [3] [4] [0]
[0] [1] [2] [3] [4]
and similarly, θ([2]) =
,
[2] [3] [4] [0] [1]
[0] [1] [2] [3] [4]
[0] [1] [2] [3] [4]
θ([3]) =
and θ([4]) =
.
[3] [4] [0] [1] [2]
[4] [0] [1] [2] [3]
21.5 (a) If [a]6 = [b]6 then 6|b−a. Since 3|6, this implies that 3|b−a, so [a]3 = [b]3 .
So θ is well defined.
(b) For any [a]6 , [b]6 ∈ Z6 we have
θ([a]6 ⊕ [b]6 ) = θ([a + b]6 ) = [a + b]3 = [a]3 ⊕ [b]3 = θ([a]6 ) ⊕ θ([b]6 )
So θ is a homomorphism Z6 → Z3 .
(c) For any [a]6 ∈ Z6 we have
[a]6 ∈ ker θ ⇐⇒ θ([a]6 ) = [0]3 ⇐⇒ [a]3 = [0]3 ⇐⇒ 3|a.
So
ker θ = {[a]6 ∈ Z6 : 3|a} = {[0]6 , [3]6 }.
b+d
21.16 Observe that for [ ac db ] ∈ M (2, Z) we have λr ([ ac db ]) = [ 10 10 ][ ac db ] = [ a+c
0
0 ].
p q
a
b
So for any [ c d ] and [ r s ] in M (2, Z),
b+q
a+p+c+r b+q+d+s
λr ([ ac db ] + [ pr qs ]) = λr ([ a+p
]
c+r d+s ]) = [
0
0
p+r q+s
p q
b+d
a b
= [ a+c
0
0 ]+[ 0
0 ] = λr ([ c d ]) + λr ([ r s ]).
1
So λr is a homomorphism. Moreover,
b+d
0 0
λr ([ ac db ]) = [ 00 00 ] ⇐⇒ [ a+c
0
0 ] = [ 0 0 ] ⇐⇒ c = −a &d = −b.
So
a b
ker λr = {[ −a
−b ] : a, b ∈ Z}.
21.17* Let θ : G → H be a homomorphism; we are to show that o(θ(a))|o(a).
If a has infinite order then this doesn’t really make sense (as o(a) hasn’t been
defined), so let’s ignore this case!
If o(a) is finite, let n = o(a). We have θ(a)n = θ(an ) = θ(e) = e, so θ(a) has
finite order, say m. We want to show that m|n. Dividing n by m, we can
write
n = mq + r for some q, r ∈ Z with 0 ≤ r < m.
Now m|mq, so θ(a)mq = e. So
e = θ(a)n = θ(a)mq+r = θ(a)mq θ(a)r = e θ(a)r = θ(a)r .
Since m is the smallest power k ≥ 1 with θ(a)k = e and 0 ≤ r < m, we can
only have r = 0. So n = mq, so m|n.
21.19 This is false in general (it is true for abelian groups, though). For example,
take G = S3 and N = h(123)i C G; then taking g = (12) and n = (123) gives
gng −1 = (12)(123)(12)−1 = (12)(123)(12) = (132) 6= (123) = n.
22.2 (a) Since [3] + [3] + [3] = [9] = [1] in Z8 , any subgroup of Z8 containing [3]
contains [1]. Since Z8 = h[1]i is generated by [1], we must have h[3]i = Z8 .
So Z8 /h[3]i = Z8 /Z8 has order 8/8 = 1 by Lagrange’s theorem.
(b)The elements of h3i/h6i are the (right) cosets of h6i in h3i, and we know
that these partition h3i.
We have h6i = h6i + 0 = {6n : n ∈ Z} and h6i + 3 = {6n + 3 : n ∈ Z}. These
cosets are not equal (since 0 is in h6i but not h6i + 3) and the union of these
two cosets is all of h3i = {3m : m ∈ Z}. Hence there are no other cosets of
h6i in h3i. So h3i/h6i = {h6i, h6i + 3} and the order of h3i/h6i is two.
22.6 Let N be a normal subgroup of a cyclic group G. Then G = hai for some
a ∈ G. If b ∈ G = hai then b = ak for some k ∈ Z, so
N b = N (ak ) = (N a)k .
So every element of G/N is a power of N a; so hN ai = G/N .
2
22.12* A word about notation: we are used to writing nq = q + q + · · · + q (n times)
rather than q n in an additive group. Since it is natural to write Q/Z as an
additive group, we will write nx = x + x + · · · + x (n times) in this group,
too. Observe that if x = Z + q for some q ∈ Q, then
nx = (Z + q) + (Z + q) + · · · + (Z + q) (n times)
= Z + (q + q + · · · + q) (by the definition of the operation on Q/Z)
= Z + nq.
Let x ∈ Q/Z. Then x = Z + m
for some n, m ∈ Z with n > 0 and (m, n) = 1
n
is
written
as
a
fraction
is in its lowest terms). Then
(that is, m
n
nx = Z + n m
= Z + m = Z = eQ/Z .
n
So x has finite order. Moreover, for k ∈ Z we have
kx = Z +
km
n
which is equal to Z if and only if km
∈ Z ⇐⇒ n|km. Since (m, n) = 1, this
n
)
=
n.
implies that n|k. So o(Z + m
n
3