Prove that the 4-group V is normal subgroup of $S_4$
First, by using the multiplication table, I am able to prove that 4-group V is subgroup of $S_4$.
But I face problem in proving that $\forall x\in S_4, xVx^{-1}=V$ since the general formula of $x$ is not given.
And this is one of the question in the book Rotman J.J Introduction to the theory of groups under the subtopic Isomorphism Theorem. So I wonder that is it possible to find a homomorphism $f:S_4\rightarrow H$ such that the kernel of $f$ is 4-group V, indirectly implying that 4-group V is subgroup of $S_4$ by First Isomorphism Theorem.
By the way, I haven't learn about conjugacy class, so if possible try to avoid using that concept to prove this.
Best Answer
Let $X$ be the set of all partitions of the set $\{1,2,3,4\}$ consisting of two parts with two elements each, so that $$X=\Bigl\{\big\{\{1,2\},\{3,4\}\big\},\big\{\{1,3\},\{2,4\}\big\},\big\{\{1,4\},\{2,3\}\big\}\Bigr\}.$$ If $g\in S_4$ and $\pi=\big\{\{a,b\},\{c,d\}\big\}\in X$, then $\big\{\{g(a),g(b)\},\{g(c),g(d)\}\big\}$ is another element of $X$, which we may denote $g\cdot\pi$. This defines a permutation $\hat g:\pi\in X\mapsto g\cdot \pi \in X$ of $X$, and it is easy to see that the map $g\in S_4\mapsto \hat g\in S(X)$, with $S(X)$ the group of all permutations of $X$, is a homomorphism of groups.
Check that it is surjective and find its kernel.