Let <G, *> be a group under the operation *. Then G has the following elementary properties.
I. Uniqueness of identity element
The identity element of a group is unique.
Proof: If possible, suppose that e1, e2 are two identity elements of a group .
∴ e1 * e2 = e2 (Since e1 is identity element) ... (1)
also e1 * e2 = e1 (Since e2 is identity element) ... (2)
Thus e1 = e2 (From (1) and (2))
∴ Thus identity element of a group is unique.
II. Uniqueness of inverse element
The inverse of each element of a group is unique.
Proof: Let e be the identity element of the group (G, *) and a ∈ G be an arbitrary element.
If possible, let b1, b2 ∈ G, be two inverses of a
∴ a * b1 = e = b1 * a (∵ b1 is inverse of a) ... (1)
and a * b2 = e = b2 * a (∵ b2 is inverse of a) ... (2)
Now b1 = b1 * e (Since e is identity of G)
= b1 * (a * b2) [∵ of (2)]
= (b1 * a) * b2 (By associativity in G)
III. Cancellation laws hold in a group
I. Uniqueness of identity element
The identity element of a group is unique.
Proof: If possible, suppose that e1, e2 are two identity elements of a group .
∴ e1 * e2 = e2 (Since e1 is identity element) ... (1)
also e1 * e2 = e1 (Since e2 is identity element) ... (2)
Thus e1 = e2 (From (1) and (2))
∴ Thus identity element of a group is unique.
II. Uniqueness of inverse element
The inverse of each element of a group is unique.
Proof: Let e be the identity element of the group (G, *) and a ∈ G be an arbitrary element.
If possible, let b1, b2 ∈ G, be two inverses of a
∴ a * b1 = e = b1 * a (∵ b1 is inverse of a) ... (1)
and a * b2 = e = b2 * a (∵ b2 is inverse of a) ... (2)
Now b1 = b1 * e (Since e is identity of G)
= b1 * (a * b2) [∵ of (2)]
= (b1 * a) * b2 (By associativity in G)
= e * b2 m [∵ of (1)]
= b2
∴ b1 = b2
Here each element of a group has unique inverse.
III. Cancellation laws hold in a group
For a, b, c ∈ G, we have
a * b = a * c ⇒ b = c (Left cancellation law)
b * a = c * a ⇒ b = c (right cancellation law)
Proof: Let a, b, c ∈ G.
Since a ∈ G so a-1∈ G such that
a-1 * a = e = a * ... (1)
Now suppose that a * b = a * c
⇒ a-1 (a * b) = a-1 (a * c)
⇒ a-1 (a * b) = a-1 (a * c) (By associative law in G)
⇒ e * b = e * c
⇒ b = c
∴ a * b = a * c ⇒ b = c
Similarly, we can prove that
b * a = c * a ⇒ b = c
IV. For every a ∈ G, (a-1)-1 = a.
Proof: ∀ a ∈ G ⇒ a-1∈ G.
then aa-1 = e = a-1a
⇒ inverse of a is a-1
Again a-1a = e = aa-1
⇒ inverse of a-1 is a
i.e., (a-1)-1 = a.
V. Reversal law for inverse of the product
(a * b)-1 = b-1 * a-1 ∀ a, b ∈ G.
Proof: Since a, b ∈ G
∴ a * b ∈ G
⇒ c ∈ G, where c = a * b ... (1)
Also b, a ∈ G
⇒ b-1 , a-1 ∈ G
⇒ b-1 * a-1 ∈ G
⇒ d ∈ G where d = b-1 * a-1 ... (2)
Consider c * d = (a * b) * d [∵ of (1)]
= a * (b * d) (By associativity in G)
= a * (b * (b-1 * a-1)) [∵ of (2)]
= a * ((b * b-1 ) * a-1)) (By associativity in G)
= a * (e * a-1)
= a * a-1 = e.
∴ c * d = e.
Now consider d * c = (b-1 * a-1) * c [∵ of (2)]
= b-1 *(a-1 * c) (By associativity in G)
= b-1 *(a-1 * (a * b)) [∵ of (2)]
= b-1 * ((a-1 * a) * b) (By associativity in G)
= b-1 * (e * b)
= b-1 * b = e
∴ d * c = e
∴ c * d = e = d * c
⇒ c-1 = d
⇒ (a * b)-1 = b-1 * a-1
VI. If a, b ∈ G be any elements. Then the equation a * x = b and y * a = b have unique solution in G.
Proof: We first prove that the equation a * x = b has a solution in G.
Since a ∈ G, so ∃ a-1∈ G such that
a * a-1 = e = a-1 * a ... (1)
Since a-1, b ∈ G so a-1* b ∈ G.
Take x = a-1* b
∴ x ∈ G
Now a * x = a * (a-1* b)
= (a * a-1) * b (By associativity in G)
= e * b = b
∴ the equation a * x = b has a solution in G.
Uniqueness: Let x1, x2 be two solutions of the equation a * x = b in G.
∴ a * x1= b and a * x2= b
⇒ a * x1= a * x2 ⇒ x1 = x2
(By left cancellation law in a group)
Hence the equation a * x = b has a unique solution in G.
Similarly, we can prove that the equation y * a = b has a unique solution in G.
VII. Left identity and right identity are the same in a group.
Proof: Let e and e' be the left identity and right identity in a group (G, *).
then,
e * e' = e' (Here e is the left identity)
also e * e' = e (Here e' is the right identity)
Thus e' = e
Hence left identity and right identity in a group are same.
VII. Left inverse and right inverse of every element in a group is same.
Proof: Let e be the identity of the group (G, *) and b and c be the left and right inverse of the element a ∈ G respectively. Then
b * a = e and a * c = e
Now b = b * e
= b * (a * c)
= (b * a) * c
= e * c
= c.
Hence the left and right inverse of every element in a group is same.
= a * (b * d) (By associativity in G)
= a * (b * (b-1 * a-1)) [∵ of (2)]
= a * ((b * b-1 ) * a-1)) (By associativity in G)
= a * (e * a-1)
= a * a-1 = e.
∴ c * d = e.
Now consider d * c = (b-1 * a-1) * c [∵ of (2)]
= b-1 *(a-1 * c) (By associativity in G)
= b-1 *(a-1 * (a * b)) [∵ of (2)]
= b-1 * ((a-1 * a) * b) (By associativity in G)
= b-1 * (e * b)
= b-1 * b = e
∴ d * c = e
∴ c * d = e = d * c
⇒ c-1 = d
⇒ (a * b)-1 = b-1 * a-1
VI. If a, b ∈ G be any elements. Then the equation a * x = b and y * a = b have unique solution in G.
Proof: We first prove that the equation a * x = b has a solution in G.
Since a ∈ G, so ∃ a-1∈ G such that
a * a-1 = e = a-1 * a ... (1)
Since a-1, b ∈ G so a-1* b ∈ G.
Take x = a-1* b
∴ x ∈ G
Now a * x = a * (a-1* b)
= (a * a-1) * b (By associativity in G)
= e * b = b
∴ the equation a * x = b has a solution in G.
Uniqueness: Let x1, x2 be two solutions of the equation a * x = b in G.
∴ a * x1= b and a * x2= b
⇒ a * x1= a * x2 ⇒ x1 = x2
(By left cancellation law in a group)
Hence the equation a * x = b has a unique solution in G.
Similarly, we can prove that the equation y * a = b has a unique solution in G.
VII. Left identity and right identity are the same in a group.
Proof: Let e and e' be the left identity and right identity in a group (G, *).
then,
e * e' = e' (Here e is the left identity)
also e * e' = e (Here e' is the right identity)
Thus e' = e
Hence left identity and right identity in a group are same.
VII. Left inverse and right inverse of every element in a group is same.
Proof: Let e be the identity of the group (G, *) and b and c be the left and right inverse of the element a ∈ G respectively. Then
b * a = e and a * c = e
Now b = b * e
= b * (a * c)
= (b * a) * c
= e * c
= c.
Hence the left and right inverse of every element in a group is same.
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