Download Definition (Binary Operation). Let G be a set. A binary operation on G

CHAPTER 2
Groups
Definition (Binary Operation). Let G be a set. A binary operation on G
is a function that assigns each ordered pair of elements of G an element of G.
Note. This condition of assigning an element of G to each ordered pair of
G is called the closure of the set G under the given binary operation.
Example. Addition, subtraction, and multiplication in Z are binary operations; division in Z is not (8 ÷ 3 62 Z).
Example. Let Zn = {0, 1, 2, . . . , n 1}, the integers modulo n. Addition
modulo n and multiplication modulo n are binary operations.
Definition (Group). Let G be a nonempty set together with a binary
operation on G (usually called multiplication) that assigns to each ordered pair
(a, b) of elements of G an element of G denoted by ab. G is a group under this
operation if
(1) The operation is associative:
(ab)c = a(bc) 8 a, b, c 2 G.
(2) There is an identity element e in G such that
ae = ea = a 8 a 2 G.
(3) For each a 2 G, there is an inverse element b 2 G such that
ab = ba = e.
Note. b is often denoted as a 1.
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28
2. GROUPS
A group G is Abelian ( or commutative) if ab = ba 8 a, b 2 G. It is non-Abelian
if there exist a, b 2 G such that ab 6= ba.
Example.
(1) Z, Q, and R are groups under + with identity 0 and inverse a for a. None
of these are groups under ⇥ since 1 is the multiplicative identity and so 0 has
no inverse in each case.
However, Q⇤ = Q\{0} and R⇤ = R\{0} are groups under ⇥, but Z⇤ = Z\{0}
is not (e.g., 3 has no inverse).
(2) {1, 1, i, i} is a group under complex multiplication. 1 and
own inverses, and i and i are inverses of each other.
1 are their
(3) The set S of positive irrational numbers along with 1 satisfy properties (1),
(2), and (3) of the definition of group under ⇥. But
p Spis not a group since ⇥
is not a binary operation on S. Closure fails (e.g., 2 2 = 2 62 S).
⇢
a b
(4) Let M =
a, b, c, d 2 R , the set of 2 ⇥ 2 matrices with
c d



a1 b1
a b
a + a2 b1 + b2
+ 2 2 = 1
.
c1 d1
c2 d2
c1 + c2 d1 + d2

0 0
M is a group with this operation.
is the identity and the inverse of
0
0


a b
a b
is
.
c d
c d
(5) Zn = {0, 1, 2, . . . , n 1} is a group under addition modulo n where, for
j > 0, n j is the inverse of j. This is the group of integers modulo n.
2. GROUPS

a b
(6) The determinant of the 2 ⇥ 2 matrix A =
is det A = ad
c d
Consider
⇢
a b
GL(2, R) =
a, b, c, d 2 R, ad bc 6= 0
c d
with multiplication



a1 b1 a2 b2
a a + b1c2 a1b2 + b1d2
= 1 2
.
c1 d1 c2 d2
c1a2 + d1c2 c1b2 + d1d2
29
bc.
Multiplication is closed in GL(2, R) since
 det(AB) = (det A)(det
 B). Associa1 0
a b
tivity is true, but messy; the identity is
; the inverse of
is
0 1
c d
 d

b
1
d
b
ad bc ad bc =
.
c
a
c
a
ad
bc
ad bc ad bc
This is the general linear group of 2 ⇥ 2 matricies over R. Since






1 2 2 3
10 13
2 3 1 2
11 16
=
and
=
,
3 4 4 5
22 29
4 5 3 4
19 28
GL(2, R) is non-Abelian.
The set of all 2 ⇥ 2 matrices over R with matrix multiplication is not a group
since matrices with 0 determinant do not have inverses.
(7) Consider Zn with multiplication modulo n. Are there multiplicative inverses? If so, we have a group.
Suppose a 2 Zn and ax mod n = 1 has a solution (i.e., a has an inverse). Then
ax = qn + 1 for some q 2 Z =) ax + n( q) = 1 =) a and n are relatively
prime by Theorem 0.2.
Now suppose a is relatively prime to n. Then, again by Theorem 0.2,
9 s, t 2 Z 3 as+nt = 1 =) as = ( t)n+1 =) as mod n = 1 =) s = a 1.
Thus we have proven Page 24 # 11:
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2. GROUPS
Lemma (Restatement of Page 24 # 11). a 2 Z has a multiplicative inverse
modulo n () a and n are relatively prime.
Definition (U (n)).
For each n > 1, define
U (n) = {x 2 Zn|x and n are relatively prime}.
U (n) will be a group under multiplication if multiplication modulo n is closed.
Lemma. If a, b 2 U (n), then ab 2 U (n).
Proof.
a, b 2 U (n) =) 9 s1, t1, s2, t2 2 Z 3 as1 + nt1 = 1 and bs2 + nt2 = 1 =)
as1 = 1 nt1 and bs2 = 1 nt2 =)(ab)(s1s2) = 1 nt1 nt2 + n2t1t2 =)
(ab)(s1s2) + n(t1 + t2 + n2t1t2) = 1.
Let s = s1s2 and t = t1 + t2 + n2t1t2. Then (ab)s + nt = 1 =) ab and n are
relatively prime =) ab 2 U (n).
⇤
So multiplication modulo n is closed in U (n), and U (n) is an Abelian group.
Example. Consider U (14) = {1, 3, 5, 9, 11, 13}.
mod14
1
3
5
9
11
13
1
1
3
5
9
11
13
3
3
9
1
13
5
11
5
5
1
11
3
13
9
9
9
13
3
11
1
5
11
11
5
13
1
9
3
13
13
11
9
5
3
1
Corollary. Z⇤n (the nonzero integers modulo n) is a group under multiplication modulo n () n is prime.
2. GROUPS
31
(8) Rn = {(a1, a2, . . . , an)|a1, a2, . . . , an 2 R} is an Abelian group under vector
addition.
(9) For (a, b, c) 2 R3, define Ta,b,c : R3 ! R3 by
Ta,b,c(x, y, z) = (x + a, y + b, z + c).
Then T = {Ta,b,c|a, b, c 2 R} is a group under function composition.
Ta,b,cTd,e,f = Ta+d,b+e,c+f .
T0,0,0 is the identity, and the inverse of Ta,b,c is T
is Abelian.
a, b, c .
This translation group
(10) Let p be prime and F 2 {Q, R, C, Zp}.
The special linear group SL(2, F ) is
⇢
a b
SL(2, F ) =
a, b, c, d 2 F, ad
c d
bc = 1 ,
where the operation is matrix multiplication
(modulo p in Zp). It is a non
a b
d
b
Abelian group. The inverse of
is
.
c d
c a

6 2
In SL(2, Z7), consider A =
. det A = 6 · 5 2 · 4 = 22 = 1 mod 7.
4 5


5
2
5 5
A 1=
=
mod 7 since
4 6
3 6




6 2 5 5
36 42
1 0
AA 1 =
=
=
mod 7 and
4 5 3 6
35 50
0 1




5
5
6
2
50
35
1 0
A 1A =
=
=
mod 7.
3 6 4 5
42 36
0 1
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2. GROUPS
GL(2, F ) is also a group under matrix multiplication (modulo p for Zp). Also,
in the case of Zp, interpret division by ad bc as multiplication by the inverse
of ad bc modulo p. GL(2, F ) is non-Abelian.

9 6
In GL(2, Z11), consider A =
. det A = 63 48 = 15 = 4 mod 11. The
8 7
inverse of 4 mod 11 is 3 mod 11 since 4 · 3 = 12 = 1 mod 11. Then



7·3
6·3
21
18
10 4
A 1=
=
=
mod 11 since
8·3 9·3
24 27
9 5




9 6 10 4
144 66
1 0
AA 1 =
=
=
mod 11 and
8 7 9 5
143 67
0 1




10
4
9
6
122
88
1 0
A 1A =
=
=
mod 11.
9 5 8 7
121 89
0 1
Theorem (2.1 — Uniqueness of the Identity).
The identity of a group G is unique.
Proof. Suppose e and e0 are identity elements of G. Then
e = ee0 = e0
from the definition of identity element, so e = e0 and the identity is unique. ⇤
Theorem (2.2 — Cancellation).
In a group G, the right and left cancellation laws hold; that is, ba = ca =)
b = c and ab = ac =) b = c.
Proof. Suppose ba = ca and let a0 be an inverse of a. Then
(ba)a0 = (ca)a0 =) b(aa0) = c(aa0) by associativity =)
be = ce =) b = c.
The proof for left cancellation is similar.
⇤
2. GROUPS
33
Theorem (2.3 — Uniqueness of Inverses).
For each element a in a group G, there exists a unique b 2 G such that
ab = ba = e.
Proof.
Suppose b and c are inverses of a. Then ab = e and ac = e =) ab = ac =)
b = c by cancellation.
⇤
Note. This allows us to ambiguously denote the inverse of g 2 G as g 1.
Notation.
g 0 = e,
gn =
ggg
· · · g}
| {z
(unambiguous by associativity).
n factors, n positive
For n < 0,
g n = (g 1)|n|, e.g., g
For every m, n 2 Z snd g 2 G,
4
= (g 1)4.
g mg n = g m+n and (g m)n = g mn.
However, in general,
(ab)n 6= anbn.
Translations to use if the group operation is “+” instead of “·.”
multiplicative
ab or a · b
e or 1
a 1
an
ab 1
additive
a+b
0
a
na
a b
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2. GROUPS
Theorem (2.4 — Socks-Shoes Property). In a group G, (ab)
Proof.
By definition and Theorem 2.3, (ab)
1
(ab)(ab)
But
1
= b 1a 1.
is the unique element in G such that
= (ab) 1(ab) = e.
(ab)(b 1a 1) = a(bb 1)a
and
1
1
= aea
1
= aa
1
=e
(b 1a 1)(ab) = b 1(a 1a)b = b 1eb = b 1b = e,
so
(ab) 1 = b 1a 1.
(In other words, b 1a 1 is the inverse of ab since it acts like an inverse, and the
inverse is unique.)
⇤
Problem (Page 56 # 25).
1
A group G is Abelian () (ab)
Proof.
= a 1b
1
8 a, b 2 G.
G is Abelian () ab = ba 8 a, b 2 G () aba
aba
1
= b () aba 1b
aba 1b
1
= e () (ab)
1
1
= bb
1
()
= a 1b 1.
1
= baa
1
()
⇤
2. GROUPS
35
Theorem (2). If G is a group and a, b 2 G, there exist unique c, d 2
G 3 ac = b and da = b (i.e., the equations ax = b and xa = b have
unique solutions in G).
Proof.
Let c = a 1b. Then ac = a(a 1b) = (aa 1)b = eb = b, so c is a solution of
ax = b.
Suppose also ac0 = b. Then
c = ec = (a 1a)c = a 1(ac) = a 1b = a 1(ac0) = (a 1a)c0 = ec0 = c0.
Thus the solution of ax = b is unique.
The proof of the second half is similar.
⇤