A question on identifying normal subgroups of $S_4$

Question

Let $H=\{e,(1 2)(3 4)\}$ and $K=\{e,(12)(34),(13)(24),(14)(23)\}$ be subgroups of $S_4$, where $e$ denotes the identity element of $S_4$. Which of the following are correct?

1. $H$ and $K$ are normal subgroups of $S_4$.
2. $H$ is normal in $K$ and $K$ is normal in $A_4$.
3. $H$ is normal in $A_4$ but not normal in $S_4$.

4. $K$ is normal in $S_4$ but $H$ is not.

   I know the condition for a subgroup $H$ to be normal in $G$ but that includes too much computation as $S_4$ has $24$ elements in all. I have studied the group of symmetries of square i.e. $D_4$ and its geometrical interpretation and I think that might be a clue to solve such type of problem relatively easy. Kindly correct me if I am wrong and provide any possible short method to do this.

Answer 1

$H$ is not normal subgroup of $S_4$. For example, you can check that $(23)(12)(34)(23) = (13)(24) \not\in H$.

$K$ is normal subgroup of $S_4$, and it named as Klein subgroup of $S_4$. To prove that $K$ is normal subgroup, you can:

• Check it directly (taking $\sigma \in S_4$, compute $\sigma \alpha \sigma^{-1}$ with $\alpha \in K$)

• Or, you can see that the conjugation in $S_n$ doesn't change the cycle structure, it follows that this subgroup is a union of conjugacy classes, and therefore is normal.

By the reason, you can see that $K$ is normal group in $A_4$.

EDIT: As @Kushal Bhuyan comment, $H$ is normal in $K$ but not in $A_4$.

Answer 2

While there are 24 permutations in $S_4$ we can break the into a few varieties.

If H or K are normal for the simple transpositions. i.e. (1,2) They will be normal for all of $S_4,$ since every element of $S_4$ is composed of transpositions.

$A_4$ has two basic varieties of elements, the disjoint transpositions and the 3-cycles e.g. (1,2,3)

Answer 3

GAVD correct answer, but $H$ is normal in $K$.

(1) and (3) are false.