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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.

Arnaud D.
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Nitin Uniyal
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  • yes..1 is not correct. – Nitin Uniyal Jun 17 '16 at 01:45
  • You should've made it clear what you are asking. When you write up claims, people, at least myself, think you are asking us to prove all those claims. I edited your post. – Emre Jun 17 '16 at 01:51
  • It's a MCQ, of CSIR-NET-JRF, so basically what OP is asking which of the options are true and which are not and why. – Kushal Bhuyan Jun 17 '16 at 01:54
  • @KushalBhuyan Check out the edit history, it wasn't clear that the question was asking for that. – Emre Jun 17 '16 at 02:01
  • @Emre..read in OP, "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". – Nitin Uniyal Jun 17 '16 at 04:55

3 Answers3

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 $A_4$.

GAVD
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1

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)

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

(1) and (3) are false.

sina
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