# Solutions for Assignment 1 Part 1 2. ```Solutions for Assignment 1
Part 1
2. (a) G = Z, a &middot; b = a − b. Not a group. The operation isn’t associative: (1 &middot; 1) &middot; 1 =
0 &middot; 1 = −1, but 1 &middot; (1 &middot; 1) = 1 &middot; 0 = 1.
(b) G = {x ∈ Z | x &gt; 0}, with multiplication as operation. Not a group. 2 doesn’t
have an inverse.
(c) Group. This is the same as the cyclic group of order 7, with ai corresponding to
i.
(d) G = { pq | p ∈ Z, q is an odd integer}, addition as operation. If p, p0 ∈ Z and
q, q 0 ∈ Z are odd integers, then
pq 0 + qp0
p p0
+ 0 =
.
q q
qq 0
Since q and q 0 are odd, qq 0 is odd, so G is closed under addition.
Addition is associative on the real numbers, so it’s associative on this set of
rational numbers.
For any a ∈ G, a + 0 = 0 + a = a, so 0 is the identity element.
For any p ∈ Z and any odd integer q, −p
∈ G and pq + −p
= −p
+ pq = 0, so G has
q
q
q
inverses. Since it satisfies all four properties, G is a group.
3. (a) Suppose that f (x) = 2x + 1 and g(x) = −3x + 2. Compute f ∗ g and g ∗ f .
f ∗ g is the function (f ∗ g)(x) = f (g(x)) = f (−3x + 2) = −6x + 5.
g ∗ f is the function (g ∗ f )(x) = g(f (x)) = g(2x + 1) = −6x − 1.
(b) Closure: If f (x) = mx+b and g(x) = m0 x+b0 , then (f ∗g)(x) = m(m0 x+b0 )+b =
mm0 x + (mb0 + b). This is an affine function, and if m, m0 6= 0, then mm0 6= 0,
and it is an element of the group.
Associativity:
((f ∗ g) ∗ h)(x) = (f ∗ g)(h(x)) = f (g(h(x)))
(f ∗ (g ∗ h))(x) = f ((g ∗ h)(x)) = f (g(h(x))),
so the operation is associative.
Identity: Let e(x) = x. Then (f ∗ e)(x) = f (e(x)) = f (x) and (e ∗ f )(x) =
e(f (x)) = f (x), so e is the identity.
Inverses: If f (x) = mx + b and m 6= 0, then f is an invertible function and
f −1 (x) = m−1 (x−b) = m−1 x−m−1 b. By the usual properties of inverse functions,
(f −1 ∗ f )(x) = (f ∗ f −1 )(x) = x,
so f −1 ∗ f = f ∗ f −1 = e.
The identity element is e, the identity function. The group isn’t abelian because
f ∗ g 6= g ∗ f when f and g are the two elements in part (a).
1
Part 2
p. 35, #2 First, we consider the case that n ≥ 0. We proceed by induction.
Base case: When n = 0, we have (ab)0 = e = a0 b0 .
Inductive hypothesis: Suppose that n ≥ 0 and that (ab)n = an bn . Then (ab)n+1 =
(ab)n ab = an bn ab (using the inductive hypothesis). Since G is abelian, bn a = abn , so
an bn ab = an abn b = an+1 bn+1 , as desired. We conclude that (ab)n = an bn for all a, b ∈ G
and all n ≥ 0.
Now suppose that n &gt; 0 and consider (ab)−n . By definition,
n
(ab)−n = (ab)−1 = (b−1 a−1 )n .
Since n &gt; 0, the argument above shows that
(b−1 a−1 )n = (b−1 )n (a−1 )n = b−n a−n .
Since G is abelian, b−n a−n = a−n b−n as desired.
p. 35, #3 Suppose that (ab)2 = a2 b2 for all a, b ∈ G. Writing out the exponents, we get
abab = aabb. By the cancellation lemma, bab = abb, and ba = ab, as desired.
#3
b
a
V
c
d
H
There are four symmetries: the identity (I), the horizontal flip (H), the vertical flip
(V ), and the 180◦ rotation through the center (R). Any symmetry f has to send a to
one of the vertices. Because the rectangle isn’t a square, once you choose a vertex to
be f (a), the images of the other vertices are determined—f (b) is the vertex adjacent
to f (a) along a long edge, f (d) is adjacent to f (a) along a short edge, and f (c) is
opposite to f (a). So there are four symmetries, one for each vertex.
The multiplication table is
◦
I
first H
V
R
I
I
H
V
R
2
then
H
H
I
R
V
V
V
R
I
H
R
R
V
H
I
Since the set consists of all the symmetries of a rectangle, it is closed under composition and inverses. Functions are associative under composition, so the operation is
associative. Finally, the identity function is the identity element of the group, so this
is a group.
#4 Let’s work by the number of elements in a subset.
First, subgroups have to be nonempty, so there are no zero-element subgroups.
Second, every subgroup contains the identity, so there is one one-element subgroup,
{I}.
A two-element subgroup contains the identity and one other element. There are three
possibilities: {I, H}, {I, V }, and {I, R}. By the table above, H 2 = V 2 = R2 = I, so
these are all closed under the operation, so they are subgroups.
A three-element subgroup must contain the identity and two other elements. There are
three possibilities: {I, H, V }, {I, V, R}, and {I, H, R}. None of these are subgroups,
because HV = R, V R = H, and HR = V – none of them are closed.
There is only one four-element subgroup, namely the whole group. This is a trivial
subgroup.
#5
1. For any a ∈ G, we have ae = ea = a by the definition of the identity element. So
Z(e) = G.
2. We need to show that Z(a) is closed under the group operation and under taking
inverses. Suppose that b, c ∈ Z(a). Then
(bc)a = b(ca) = b(ac) = (ba)c = (ab)c = a(bc),
so bc ∈ Z(a). Furthermore, we can multiply on the left and right by b−1 to get:
ba = ab
b &middot; ba = b−1
b−1 ba &middot; b−1 = b−1 ab &middot; b−1
ab−1 = b−1 a.
−1
So b−1 ∈ Z(a). By the lemma we proved in class, this implies that Z(a) is a
subgroup of G.
3
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