group of transformations

A group of transformations on set ${S}$ is a set ${G}$.${G}$ is a set of bijections from ${S}$ to itself. And ${G}$ is a group,which means that it satisfies three properties:

Identity:${\forall a\in G}$,${ \exists e\in G}$ such that ${a\circ e=e\circ a =a}$.(${e}$ represents the identity mapping.)
Inverse:${\forall a\in G}$,${\exists a^{-1}\in G}$ such that ${a^{-1}\circ a=a\circ a^{-1}=e}$.(${a^{-1}}$ represents the inverse map of ${a}$.)
combinative :${\forall a,b,c\in G}$,we have ${(a\circ b)\circ c=a\circ (b\circ c)}$.(There is a prerequisite for this property,that is,${\forall a,b\in G,a\circ b\in G}$.The combinative property can be deduced from this prerequisite,so it is better to replace the combinative property by this prerequisite.But we have to be sure that if ${a\circ b\in G}$,${a\circ b}$ should satisfy property 1 and 2,this is easy to verify. )

Now I want to talk about another point:

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All the bijections from set ${S}$ to ${S}$ form a group of transformation.But a group of transformation is not necessarily consists of all the bijections from set ${S}$ to ${S}$.

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The first statement of this point can be verified as follows:

1.It is easy to verify that if $a$ is a bijection from $S$ to $S$,then $a^{-1}$ is also a bijection from $S$ to $S$.

2.It is easy to verify that for all bijection $a$ from $S$ to $S$,there exists an identity mapping $e$ from $S$ to $S$ such that $a\circ e=e\circ a=a$.

3.It is easy to verify that if $a,b$ is two bijections from $S$ to $S$,then $a\circ b$ is also a bijection from $S$ to $S$.And,$a\circ b$ has an inverse $b^{-1}\circ a^{-1}$,this inverse is also a bijection form $S$ to $S$ .And,$(a\circ b )\circ e=e\circ (a\circ b)=a\circ b$.

As for the second statement,I show you some examples:

Example 1.G is the set of identity mapping of set ${A}$.

Example 2.Let $S$ be the set of all the real numbers,while the transformations considered to have the form $f(x)=ax+b$.Consider the following cases,in some cases ,$G$ is a group,while in the rest of the cases,$G$ is not a group:

(1) $G=\{f(x)|a=1,b$ is an odd number$\}$.In this case,$G$ is not a group,because if it is a group,it should have an identity element for every member in it.But for $f(x)=x+1$,its indentity is $g(x)=x+0$,but 0 is not an odd number.

(2)$G=\{f(x)|a=1,b$is a positive integer or 0$\}$.In this case,$G$ is not a group,because it does't have an inverse for every member in it.For example,$f(x)=x+3$.Its inverse is $g(x)=x-3$.But -3 is a negative number.

(3)$G=\{f(x)|a=1,b$is an even number.$\}$.In this case,$G$ is a group,because first,all the elements in $G$ are bijections from $S$ to $S$.And,every element in $G$ has an inverse which is also in $G$,for example,the inverse of $f(x)=x+2$ is $g(x)=x-2$,-2 is an even number.And,every element in $G$ has an indentity element which is also in $G$ ,as 0 is an even number.And, it is easy to verify that the composition of any two elements in $G$ is also in $G$,as the sum of two even numbers is also an even number.