I have shown that in strait forward way .
gVg^(-1) = V
But is there some other way to show that ?
Sylow dint worked for me here .
Thanks for help .
If you already know that two permutations in $\displaystyle S_n$ are conjugate iff they have the very same cycle decomposition, then it is straightforward that $\displaystyle V$ is a normal subgroup of $\displaystyle S_4$ and, thus, also of $\displaystyle A_4$ since $\displaystyle V\subset A_4$ .
Tonio
Well, for example $\displaystyle (12)(345), (14)(235)$ are conjugate permutations in $\displaystyle S_n\,,\,n\ge 5$ , since they both are the product of a 2-cycle and a 3-cycle, but none of them is conjugate to $\displaystyle (13)(24)$ since this last is the product of two 2-cycles (transpositions), and etc.
Of course, one must know first that ANY permutation is expressable as the product of DISJOINT cycles...
Tonio
If you want a geometric interpretation, use the fact that A4 is the group of rotations of a regular tetrahedron. The tetrahedron has three axes, namely the lines joining the midpoints of opposite edges. A rotation of the tetrahedron takes axes to axes, and thus induces a homomorphism from A4 to S3, whose kernel is the Klein group (consisting of the identity map together with rotations through 180º about each axis).