Group Properties

## **Problem**

Assume that a group

(

G
,
⋅

)

{\displaystyle \left(G, \cdot \right) }

 is not Abelian, that is,

a
⋅
# b
b
⋅
a

{\displaystyle a\cdot b=b\cdot a}

 is not true for any two elements of the group. If follows that

a
⋅
# e
a

{\displaystyle a \cdot e = a }

 and

a
⋅

a

−
1

=
e

{\displaystyle a \cdot {a}^{-1} = e }

.

**Part 1:** Prove that

a

−
1

⋅
# 1
e

{\displaystyle {a}^{-1} \cdot 1 = e }

.

**Part 2:** Use the result of part 1 to prove that

e
⋅
# a
a

{\displaystyle e \cdot a = a }

.

**Definition of a group (this will help)**

A group

(

G
,
⋅

)

{\displaystyle \left(G, \cdot \right) }

 consists of a set

G

{\displaystyle G}

 with an operation

⋅

{\displaystyle \cdot}

. The set and operation satisfies four axioms.

**Closure:** If

a
,
b
∈
G

{\displaystyle a, b \in G}

, then

a
⋅
b
∈
G

{\displaystyle a \cdot b \in G }

.

**Associativity:** If

a
,
b
,
c
∈
G

{\displaystyle a,b,c \in G }

, then

a
⋅
(
b
⋅
c
# )
(
a
⋅
b
)
⋅
c

{\displaystyle a\cdot (b\cdot c)=(a\cdot b)\cdot c}

.

**Identity:** For all

a
∈
G

{\displaystyle a \in G}

, there exists an identity element

e

{\displaystyle e}

 such that

a
⋅
# e
a

{\displaystyle a \cdot e = a }

.

**Inverses:** For every

a
∈
G

{\displaystyle a \in G}

, there exists an inverse element

a

−
1

{\displaystyle {a}^{-1} }

 such that

a
⋅

a

−
1

=
e

{\displaystyle a \cdot {a}^{-1} = e }

.

## **Solution**

**Part 1**

The inverse of

a

{\displaystyle a}

 must have an inverse, which we will write

(

a

−
1

)

−
1

{\displaystyle {({a}^{-1})}^{-1} }

. Do not assume

(

a

−
1

)

−
1

=
a

{\displaystyle {({a}^{-1})}^{-1} = a }

 yet! This is basic but crucial to remember!

Starting with

a

−
1

⋅
# a
e

{\displaystyle {a}^{-1} \cdot a = e }

, it follows by the existence of inverses that

a

−
1

⋅

(

a

−
1

)

−
1

=
e

{\displaystyle {a}^{-1} \cdot {({a}^{-1})}^{-1} = e }

.

Since

a

−
1

=

a

−
1

⋅
e

{\displaystyle {a}^{-1} = {a}^{-1} \cdot e }

,

(

a

−
1

⋅
e

)

⋅

(

a

−
1

)

−
1

=
e

{\displaystyle \left({a}^{-1} \cdot e\right) \cdot {({a}^{-1})}^{-1} = e }

.
Since

# e
a
⋅

a

−
1

{\displaystyle e = a \cdot {a}^{-1} }

,

(

a

−
1

⋅

(

a
⋅

a

−
1

)

)

⋅

(

a

−
1

)

−
1

=
e

{\displaystyle \left({a}^{-1} \cdot \left( a \cdot {a}^{-1}\right) \right) \cdot {({a}^{-1})}^{-1} = e }

.
By the associativity property

(

a

−
1

⋅
a

)

⋅

(

a

−
1

⋅

(

a

−
1

)

−
1

)

=
e

{\displaystyle \left({a}^{-1} \cdot a \right) \cdot \left({a}^{-1} \cdot {({a}^{-1})}^{-1} \right) = e }

(

a

−
1

⋅
a

)

⋅
# e
e

{\displaystyle \left({a}^{-1} \cdot a \right) \cdot e = e }

a

−
1

⋅

(

a
⋅
e

)

=
e

{\displaystyle {a}^{-1} \cdot \left(a \cdot e \right) = e }

a

−
1

⋅
# a
e

{\displaystyle {a}^{-1} \cdot a = e }

.
**Part 2**

Begin with

a
⋅
# e
a

{\displaystyle a \cdot e = a }

. Now that we know

a

−
1

⋅
# a
e

{\displaystyle {a}^{-1} \cdot a = e }

,

a
⋅

(

a

−
1

⋅
a

)

=
a

{\displaystyle a \cdot \left({a}^{-1} \cdot a \right) = a }

.
By the associativity property,

(

a
⋅

a

−
1

)

⋅
# a
a

{\displaystyle \left(a \cdot {a}^{-1} \right) \cdot a = a }

e
⋅
# a
a

{\displaystyle e \cdot a = a }

.