Math337 ? Group Theory Semester 1, 2004 Assignment 1A Solutions

Assignment 1A Solutions
Math337 — Group Theory
Semester 1, 2004
1. Which of the following subsets of Z are groups under addition? Give brief reasons for your
answers.
(a) A = the set of even integers;
(b) B = the set of odd integers;
(c) C = the set of non-negative integers;
(d) D = {0}
(e) E = the set of integers which are expressible as 42m + 1023n for integers m, n.
Solution:
(a) A is a subgroup of Z, and so is a group. 0 ∈ A, soAis non-empty. If 2n and 2m ∈ A,
then 2n − 2m =2(n−m) ∈ A.
(b) B is not a group because 1 + 1 /∈ B.
(c) C is not a group. If it were, 0 would have to be the identity because 0 + x = x +0=x
for all x ∈ C. However, the ‘equation’ 1 + x = 0 has no solution in C.
(d) D is clearly a group. 0 is the identity, its own inverse and the addition is obviously
associative.
(e) E is a subgroup of Z, and so is a group. 42 = 42(1) + 1024(0) ∈ E, soE is non-empty.
If 42m1 + 1023n1 and 42m2 + 1023n2 ∈ E, then (42m1 + 1023n1) − (42m2 + 1023n2) =
42(m1 − m2) + 1023(n1 − n2) ∈ E.
2. Which of the following subsets of C are groups under multiplication? Give brief reasons for
your answers.
(a) A = the set of non-zero rational numbers;
(b) B = the set of positive integers;
(c) C = {1, −1,i,−i};
(d) D = {1, √ 2, 2};
(e) E = a + bi|a >0;
(f) F = {1,π,π 2 ,π 3 ,...}.
Solution:
(a) A is a subgroup of C # , and so is a group. 1 ∈ A, soAis non-empty. If n1
m1 −1 ∈ A.
then n1
m1
n2
m2
= n1m2
n2m1
and n2
m2
(b) B is not a group. If it were, 1 would be the identity because 1x = x(1) = x for all x ∈ B.
However, the ‘equation’ 2x = 1 has no solution in B.
(c) C is a subgroup of C # , and so is a group. It is non-empty, closed under multiplication
(16 things to check) and taking inverses (4 things to check) and inherits associativity
from C # .
(d) D is not a group. 2(2) /∈ D.
(e) E is not a group. (1 + i)(1+2i) /∈ E.
(f) F is not a group. If it were, 1 would be the identity because 1x = x for all x ∈ B.
However, the ‘equation’ πx = 1 has no solution in F .
⎛
1
3. Consider the set of all matrices over Z2 of the form ⎝ 0
a
1
⎞
b
c ⎠. There are 8 matrices in this
0
set as there are 2 choices for each of a, b and c.
0 1
(a) Show that this set is a group of order 8 under matrix multiplication. Does it satisfy the
commutative law?
∈ A,

(b) Find all the elements of order 4 in the above group.
(c) Show that the above group is a dihedral group of order 8, that is find an element A of
order 4 and an element of order 2 such that BA = A−1B. Solution:
(a) There are 8 elements of this set, corresponding to the 8 possible choices of (a, b, c) ∈ Z3 2.
Clearly
⎛
1
⎝ 0
0
1
⎞ ⎛
0 1
0⎠⎝0
a
1
⎞ ⎛
b 1
c ⎠ = ⎝ 0
a
1
⎞ ⎛
b 1
c ⎠ ⎝ 0
0
1
⎞ ⎛
0 1
0⎠
= ⎝ 0
a
1
⎞
b
c ⎠ ,
0 0 1 0 0 1 0 0 1 0 0 1 0 0 1
⎛
1
so ⎝ 0
0
1
⎞
0
0⎠is
the identity. It is easy to check that
0 0 1
⎛
1
⎝ 0
a
1
⎞ ⎛
b 1
c ⎠ ⎝ 0
a
1
⎞ ⎛
b+ ac 1
c ⎠ = ⎝ 0
a
1
⎞ ⎛
b+ ac 1
c ⎠ ⎝ 0
a
1
⎞ ⎛
b 1
c ⎠ = ⎝ 0
0
1
⎞
0
0⎠
0 0 1 0 0 1 0 0 1 0 0 1 0 0 1
so inverses exist, and multiplication of⎛matrices ⎞is
⎛always
associative, ⎞ ⎛ so the ⎞set
is a
1 1 0 1 0 0 1 1 1
group. It is not commutative because ⎝ 0 1 0⎠⎝0
1 1⎠
= ⎝ 0 1 1⎠while
⎛
1
⎝ 0
0
1
⎞ ⎛
0 1
1⎠⎝0
1
1
⎞ ⎛
0 1
0⎠
= ⎝ 0
1
1
⎞
0
0
1⎠.
0 1 0 0 1 0 0 1
0 0 1
⎛
1
(b) Since ⎝ 0
a
1
0 0 1
⎞2
⎛
b 1
c ⎠ = ⎝ 0
0
1
0 0 1
⎞
ac
0 ⎠, it is clear that all the elements of the group have
0 0 1 0 0
⎛
1
1
order less than 2 apart from ⎝ 0
1
1
⎞ ⎛
0 1
1⎠and
⎝ 0
1
1
⎞
1
1⎠.
The square of these two
⎛
1
elements is ⎝ 0
0
1
⎞
0 0 1 0 0 1
1
0⎠which
has order 2, so they must have order 4.
⎛
1
(c) Let A = ⎝ 0
0
1
1
0
⎞
1
⎛
0
1
1⎠,
and B = ⎝ 0
1
1
⎞
0
0⎠,
soAhas order 4 and B has order 2. It is
0 0 1
⎛
1
easy to check that BAB = ⎝ 0
0 0
⎞
1
1 1
1 1⎠
= A
0 0 1
−1 . [There are two possible choices for A,
my A and my A −1 . There are 4 choices for B — any of the elements of order 2 apart
from A 2 .]
4. The group table for the a finite group displays the results of the binary operation ∗. The entry
in the row labelled g and column labelled h is the element g ∗ h.
(a) Prove that in the group table of a group of order n, every element appears exactly once
in every row and column.
(b) The following is a partially completed group table for a group. Complete it.
a b c d
a a
b c
c a
d d

Which element is the identity?
(c) G = {1,a,b} is a group of order 3. Complete its group table.
Solution:
(a) Suppose g ∗ h1 = g ∗ h2. Thenh1 = h2 by left cancellation. So noentry can occur twice
in the row for g. Suppose h ∈ G. Then g ∗ (g −1 ∗ h) =(g ∗ (g −1 ) ∗ h = h, soh does
appear in the row for g. That is, each element of the group occurs precisely once in each
row.
Similarly h1 ∗ g = h2 ∗ g implies that h1 = h2, so no entry can occur twice in the column
for g, and (h ∗ g −1 ) ∗ g = h, so every element does occur in the column for g.
(b) Clearly b is the identity, which allows us to complete the row and column for b. I’ve
added these entries in red. Then d must occur in the column for c and cannot be in the
row for d, so we must have ac = d and dc = b. Similarly d must occur in the row for
c and cannot be in the column for d, sowemusthaveca = d and cd = b. I’ve added
these entries in green . Then ad = c, since it can’t be a, and the remaining entries in
blue follow immediately.
a b c d
a b a d c
b a b c d
c d c a b
d c d b a
(c) If 1 is the identity, then the other 2 elements must have order 3, so a 2 =1ora, sowe
must have a 2 = b. There is then only one way to fill out the rest of the table by part (a).
1 a b
1 1 a b
a a b 1
b b 1 a