Find some permutation group(s) isomorphic to Klein's 4-group** 
Find some permutation group(s) isomorphic to  Klein's 4-group

Klein's 4-group $=V=\{e,a,b,ab\}$. such that order of every non-identity element of V is 2. Also V is abelian.
So it is clear that the permutation group(s) we have to find should be order 4, abelian and ....
I don't know what other things that are kept to be preserved while construct the permutation group(s)?
Please help me to solve it? Please don't only give the examples, please also supply me the reason behind your construction.
Please help.
 A: Consider the set of permutations $T_\alpha : V \mapsto V$ defined by $T_\alpha(x)=\alpha x$.
Verify that this set of permutations $S$ together with the operation function composition is a group and also  which is isomorphic to $V$.
This is nothing but the Cayley's theorem

Claim is that $S=\{T_e,T_a,T_b,T_{ab}\}$ is a group isomorphic to $V$. Also see that $T_{xy}=T_xT_y$, using this compute all possible combinations and you will find it is closed and abelian. Just apply the definition to get $T_e^2=T_a^2=T_b^2=T_{ab}^2=T_e$.
Now identify the isomorphism $\phi:G\mapsto S$ such that $\phi(x)=T_x.$ Verify that $\phi$ is an ismomorphism i.e show that $\phi$ is a bijection and $\phi(xy)=\phi(x)\phi(y)$. Also don't forget to verify that $\phi$ is indeed a function.
A: If you knew Cayley's theorem you could solve your problem by simply applying the proof of Cayley's theorem to the group $V.$ I assume you don't know that theorem.
In the group $V$ the elements $a,b$ are elements of order two which commute, so let's start by finding two commuting permutations of order two. Permutations of order two are products of one or more disjoint transpositions, such as $(1\ 2),\ (1\ 2)(3\ 4),\ (1\ 2)(3\ 4)(5\ 6),$ and like that. Multiplication of permutations is usually noncommutative, e.g. $(1\ 2)(2\ 3)=(1\ 2\ 3)\ne(1\ 3\ 2)=(2\ 3)(1\ 2),$ but multiplication of disjoint permutations is commutative. So let's try taking two disjoint transpositions: $a=(1\ 2),\ b=(3\ 4),\ ab=ba=(1\ 2)(3\ 4),$ and let's hope that
$$G=\{(1),\ (1\ 2),\ (3\ 4),\ (1\ 2)(3\ 4)\}$$
is a group of permutations, and that it's isomorphic to $V.$
Is it?
A: Beware of presentations of groups as permutation groups. Every finite group has its natural representation because the (left-) multiplication on the other elements acts as a permutation. In the case of the Klein group with elements $e, u, v, w$, with identity $e$ and $w = uv$. $e$ gives $()$, $u$ gives the premutation $(e,u)(v,w)$, $v$ gives $(e,v)(u,w)$ and $w$ gives $(e,w)(u,v)$, so that Klein group has another permutation presentation as $$K = \{(e), (e,u)(v,w), (e,v)(u,w),(e,w)(u,v)\} $$. The main difference with the permutation presentation in the first answer is that here all the permutations are even.
A: It's well known that  Klein's four group embeds in $S_4$.
Actually,  $S_4$ contains four Klein four subgroups, only one of which is normal.
They are: $\{e,(12), (34),(12)(34)\},\{e,(13),(24),(13)(24)\},\{e,(14),(23),(14)(23)\} $ and  $\{e,(12)(34),(13)(24),(14)(23)\} $.
The last one is the normal one, because conjugation preserves cycle structure.
