# 2.2: Properties of Groups

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##### 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)

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.

This page titled 2.2: Properties of Groups is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.