- 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. )
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}$.
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.