2.2: Properties of Group Elements

Theorem \(\PageIndex{1}\)

Let \((G, \ast)\) be a group.    Then

1. The identity is unique,

2. For each \(a \in G\), one and only one inverse exists.

3. For each \(a \in G, \ (a^{-1})^{-1}=a\).

4. \((a\ast b)^{-1}=b^{-1} \ast a^{-1}, \forall a,b \in G.\)

Note if \(G\) is abelian,\((a\ast b)^{-1}=b^{-1} \ast a^{-1}=a^{-1} \ast b^{-1}, \forall a,b \in G\).

5. If \(xy=xz, \forall x,y,z \in G\), then  \(y=z\).  (Left cancellation)

6. If \(yx=zx, \forall x,y,z \in G\), then  \(y=z\).  (Right cancellation)

 

Answer

Proof:

Let \((G,\ast)\) be a group.

1. We shall show that identity is unique.

Assume that \(G\) has two identity elements, \(e_1\) and \(e_2\).

Thus \(a \ast e_1=e_1 \ast a=a\) and \(a \ast e_2=e_2 \ast a=a\), \(\forall a \in G\).

We will show that \(e_1=e_2\).

Consider \(e_1 \ast e_2=e_2 \ast e_1 =e_1\) and \(e_2 \ast e_1=e_1 \ast e_2 =e_2\).

Thus since \(e_1=e_2\), the identity is unique.◻

 

2. We shall show that for each \(a \in G\), one and only one inverse exists.

Let \(a \in G\).

Assume that \(a\) has two inverses, \(b\) and \(c\).

Then \(a \ast b=b \ast a=e\) and \(a \ast c=c \ast a=e\).

We shall show that \(b=c\).

Consider \(b=b \ast e\)

      \(=b \ast(a \ast c)\)

      \(=(b \ast a) \ast c\)

      \(=e \ast c\)

      \(=c\).

Thus \(b=c\) and for each \(a \in G\), there exists one and only one inverse.

3. We shall show that for each \(a \in G, \ (a^{-1})^{-1}=a\).

Let \(a \in G\). Then  \(a \ast a^{-1}=a^{-1}\ast a=e\).

Thus \((a^{-1})^{-1}=b^{-1}=a\).

Assume that \(G\) has two identity elements, \(e_1\) and \(e_2\).

Thus \(a \ast e_1=e_1 \ast a=a\) and \(a \ast e_2=e_2 \ast a=a\), \(\forall a \in G\).

We will show that \(e_1=e_2\).

Consider \(e_1 \ast e_2=e_2 \ast e_1 =e_1\) and \(e_2 \ast e_1=e_1 \ast e_2 =e_2\).

Thus since \(e_1=e_2\), the identity is unique.◻

4. 

Let \(a,b \in G\).

Then \(a \ast b \in G\), \(a^{-1} \in G\), \(b^{-1} \in G\) and \(a \ast b \in G\).

Consider \((b^{-1} \ast a^{-1})(a \ast b)= b^{-1} \ast (a^{-1}a) \ast b)\)

        \(=b^{-1}\ast e \ast b\)

        \(=b^{-1}\ast b\)

        \(=e\).

And consider  \((a \ast b)(b^{-1} \ast a^{-1})=a \ast ( b^{-1} \ast b) \ast a^{-1}\)

   \(=a^{-1}\ast e \ast a\)

           \(=a^{-1}\ast a\)

   \(=e\).

Since \((b^{-1} \ast a^{-1})(a \ast b)=e= (a \ast b)(b^{-1} \ast a^{-1})\), then \((a\ast b)^{-1}=b^{-1} \ast a^{-1}, \forall a,b \in G\).◻

5.

Let \(x,y,z \in G\) s.t. \(xy=xz\).

Note:  \(x^{-1} \in G\) since \(G\) is a group.

Thus \(x^{-1}(xy)=x^{-1}(xz)\)

Thus \((x^{-1}x)y=(x^{-1}x)z\) since  associative.

Thus \(ey=ez\) and \(y=z\).◻

6. 

Let \(x,y,z \in G\) s.t. \(y \ast x=z \ast x\).

Note:  \(x^{-1} \in G\) since \(G\) is a group.

Thus \((y \ast x) \ast x^{-1}=(z \ast x) \ast x^{-1}\)

Thus \(y \ast (x^{-1} \ast x)=z \ast (x^{-1} \ast x)\) since  associative.

Thus \(y \ast e=z \ast e\) and \(y=z\).◻

 

Theorem \(\PageIndex{2}\)

Let \(G\) be a group.

Let \(g,h \in G\) and \(m,n \in \mathbb{N}\).

Then

1. \(g^m \ast g^n=g^{m+n}\) where \(g^m=g \ast \cdots \ast g\).  Note there would be \(m\) \(g\)’s.

2. \((g^m)^n=g^{mn}\).

3. \((gh)^{-m}=(h^{-1} g^{-1})^m\).   

Note

With the addition operation \(g^m=mg.\) 

Example \(\PageIndex{1}\)

Let \(G\) be a group and suppose that \((ab)^2=a^2b^2\) for all \(a\) and \(b\) in \(G\).  Prove that \(G\) is an abelian group.

Solution

Let \(G\) be a group.

Let \((ab)^2=a^2b^2, \; \forall a,b \in G\).

We shall show that \(G\) is abelian.

Let \(a,b \in G\).

We shall show that \(ab=ba\).

Note that \((a \star b)\star (a \star b)= (a \star a) \star (b \star b)\).

Consider that \(a^{-1} \star (a \star b) \star (a \star b)\star b^{-1}=a^{-1} \star (a \star a) \star (b \star b) \star b^{-1}\).

Then \((a^{-1} \star a) \star b \star a \star (b \star b^{-1})=(a^{-1} \star a)\star a \star b \star (b \star b^{-1})\).

Thus \(e \star b \star a \star e=e \star a \star b \star e\).

Thus \(b \star a=a \star b\).◻

Note:  Operator * included for students.