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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.

Martin Sleziak
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Wang Kah Lun
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  • I'm not saying this is the best way, but you can use the multiplication table to verify that conjugating each element of $V$ by the group yields another element of $V$. – Kyle Miller Jul 23 '15 at 05:34
  • Yes. $V$ is the kernel of a surjective homomorphism from $S_4$ to $S_3$. I describe that homomorphism in this answer. Do study that but be warned that the description of that homomorphism essentially uses the concept of group action. I tried to hide it, but I'm not sure it completely can. It is also possible to give a geometric description of that realizing $S_4$ as a group of symmetries of a cube, but that is also about group actions. Well, conjugacy classes are also an instance of group actions :-) – Jyrki Lahtonen Jul 23 '15 at 06:44

4 Answers4

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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.

Batominovski
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Mariano Suárez-Álvarez
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Since $S_4$ is generated by $u=(12)$ and $v=(1234)$, it is enough to check that $uVu^{-1}=V$ and $vVv^{-1}=V$.

That solves your problem about «nor having a general formula for $x$».

Mariano Suárez-Álvarez
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Note: I am proving in first way, Not by Isomorphism theorem,

The only proper non-trivial normal subgroup of $S_4$ are $$V_4 = \{e,(12) (34) , (13)(24), (14) (23)\}$$ and $A_4$. Suppose that $N$ is a normal proper non-trivial subgroup of $S_4$. First note that $N$ does not contain a transposition, because if one transposition $τ$ lies in $N$, then $N$ contains all transpositions, hence $N = S_4$. If $N$ contains a $3$-cycle,then $N$ contains all $3$-cycles (as they are all conjugate). Therefore $N = A_4$ (as $3$-cycles generate $A_4$). If $N$ contains a $4$-cycle $(abcd)$, then it also contains its conjugate $(bacd)$, and the product $$(bacd) (abcd) = (bdc)$$. If $N$ contains a $3$-cycle, we already have that it contains $A_4$. But $N$ also contains an odd permutation. Hence $N = S_4$. Finally, if $N$ does not contain a transposition, $3$-cycle or $4$-cycle, it must contain a disjoint product of two transpositions $(ab) (cd)$. Then $N = V_4$.

Chiranjeev_Kumar
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Note that every element of $S_4$ is a product of transpositions $(1\;2)$, $(2\;3)$, or $(3\;4)$. Consider the surjective map $\varphi:S_4\to S_3$ sending all $\sigma \in S_4$ to an element of $S_3$ by switching $(3\;4)$ to $(1\;2)$ in any decomposition into a product of transpositions of $\sigma$. For example, $$\varphi\big((2\;4)\big)=\varphi\big((2\;3)(3\;4)(2\;3)\big)=(2\;3)(1\;2)(2\;3)=(1\;3)\,.$$ However, to show that $\varphi$ is a well defined group homomorphism with kernel $V$, we have to rely on the fact that $V$ is normal in $S_4$. My post is not an answer to your question, but I only wanted to point out that you can concretely construct such a homomorphism.

Batominovski
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