Page 32 - DMTH403_ABSTRACT_ALGEBRA

Unit 2: Groups

2.2 Group Notes

After understanding the concept of binary operations. Let us start defining group.

Definition: Let G be a non-empty set and * be a binary operation on G. We say that the pair
(G, * ) is a group if
G 1) * is associative:’

G 2) G contains an identity element e for * , and
G 3) every element in G has an inverse in G with respect to *.
We will now give some examples of groups.

Example: Show that (Z, +) is a group, but (Z,.) is not.
Solution: + is an associative binary operation on Z. The identity element with respect to + is 0,
and the inverse of any n  Z is (–n). Thus, (Z, +) satisfies GI, G2 md G3. Therefore, it is a group.
Now, multiplication in Z is associative and 1  Z is the multiplicative identity. But does every
element in Z have a multiplicative inverse? No. For instance, 0 and 2 have no inverses with
respect to ‘.’ Therefore, (Z,.) is not a group.
Note that (Z,.) is a semigroup since it satisfies GI. So, there exist semigroups that aren’t groups!
Actually, to show that (G, *) is a group it is sufficient to show that * satisfies the following
axioms.
G 1’) * is associative.

G 2’)  e  G such that a * e = a  a  G .
G 3’) Given a  G, 3 b  G such that a * b = e.
What we are saying is that the two sets of axioms are equivalent. The difference between them
is the following:
In the first set we need to prove that e is a two-sided identity and that the inverse b of any a  G
satisfies a * b = e and b * a = e. In the second set we only need to prove that e is a one-sided identity
and that the inverse b of any a  G only satisfies a * b = e.
In fact, these axioms are also equivalent to
G 1”) * is associative.

G 2") 3 e  G such that i * a = a  a  G.
G 3") Given a  G, 3 b  G such that b * a = e.

Clearly, if * satisfies GI, G2 and G3, then it also satisfies Gl’, G2' and G3'. The following theorem
tells us that if * satisfies the second set of axioms, then it satisfies the first set too.

Theorem 3: Let (G, * ) satisfy Gl’, G2’ and G3’. Then e * a = a  a  G. Also, given a  G, if  bG
such that a * b = e, then b * a = e. Thus, (G, *) satisfies G1, G2 and G3.
To prove this theorem, we need the following result.

Lemma 1: Let (G, * ) satisfy Gl’, G2' and G3'. If  a  G such that a *a = a, then a = e.
Proof: By G3' we know that 3 b  G such that a * b = e.

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