How to show that $K\leq S_4$ is a normal subgroup?

Question

Let $K=\{(1),(1,2),(3,4),(2,4),(1,4),(2,3)\}$ is a subgrope of $S_4$, we want to show that it is a normal subgroup.

I know that i have to examine the condition of normality in $S_4$ i.e. for all $g\in S_4, g^{-1}Kg=K$. I think this takes times or I am missing something near. Please give me the right hint. Thank you.

Edit: $K = \{(1),(12)(34),(13)(24),(14)(23)\}.$

What is your text book? Most probably it is a typo or you simply miscopied... It could probably be $,{(1), (12)(34), (13)(24), (14)(23)}\triangleleft S_4,$ — DonAntonio, Feb 23 '13 at 16:56

score 7 · Answer 1 · answered Feb 23 '13 at 16:54

Maybe your course instructor intended $K = \{(1),(12)(34),(13)(24),(14)(23)\}.$ That is, $K$ consists of the identity, together with all products of two disjoint $2$-cycles. This is a subgroup. It is also normal because the identity is conjugate only to the identity, and because the set of products of two disjoint $2$-cycles is closed under conjugacy, so that $\sigma K\sigma^{-1} = K$ for all $\sigma \in S_{4}.$

score 4 · Accepted Answer · answered Feb 23 '13 at 15:58

This isn't going to work on two fronts:

First your subset $K$ is not a subgroup: $(1 \ 2)(2 \ 3) = (1 \ 2 \ 3)$ so $K$ is not closed under the group operation.

Second your subset $K$ is not closed under conjugation: $(2 \ 3)^{-1}(1 \ 2)(2 \ 3) = (1 \ 3)$.

Maybe you meant the subgroup generated by $K$? But $K$ generates all of $S_4$ so in that case it is trivially normal.

How to show that $K\leq S_4$ is a normal subgroup?

2 Answers2

Linked

Related