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Prove that $\{id, (12)(34), (13)(24), (14)(23)\}$ is a normal subgroup of $A_4$. I see that it's possible to prove this with the following theorem: $$N \trianglelefteq G \text{ if } N\leq G \text{ and } \forall n\in N, g\in G: g^{-1}ng \in N.$$ But is there an easier way to do this? This way isn't hard but it takes quite some time to work through.

dxtr
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  • This could be proven using Sylow Theorems and knowledge about the elements of $A_4$. Have you learned about the Sylow Theorems? – ndhanson3 Dec 15 '20 at 07:56
  • That might be a little heavy-handed. Another way is with knowledge of the conjugacy classes of $S_n$ and $A_n$. – ndhanson3 Dec 15 '20 at 08:04

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It is a clear that the set is a subgroup. This requires minimal calculations. What you could do next is to calculate the conjugacy class of $(12)(34)$ in $A_4$. Almost hands-down (conjugation preserves cycle structures!!) you will see that this is $Cl_{A_4}((12)(34))=\{(12)(34),(13)(24),(14)(23)\}$. Hence your subgroup is the disjoint union of two conjugacy classes: $\{(1)\} \cup \{(12)(34),(13)(24),(14)(23)\}$ whence normal. Note that the subgroup is isomorphic to $V_4$, Klein's $4$-group. It is also equal to the commutator subgroup $[A_4,A_4]$ of $A_4$, which yields another way of showing normality.

So what I used is the fact that if $N$ is a subgroup of $G$, then $N \unlhd G$ if and only if $N=\bigcup_{n \in N}Cl_G(n)$, which you might try to prove yourself. Note that in general, unions of $G$-conjugacy classes form a normal set (meaning: closed under conjugation), but do not need to form a subgroup!

Nicky Hekster
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