Prove that $A_5$ cannot have a normal subgroup of order $2$

My Attempt:

Suppose that $H$ is a normal subgroup of $A_5$ and $|H|=2$ .

It can be shown that $H$ is contained in the center of A 5 . So elements of $H$ commutes with every element of $A_5$ .

Say $(ab)(cd) \in H$ . Since every element of $H$ is in the center , $(ab)(cd)$ commutes with every element of $A_5$ .

But we can check that $(abc)(ab)(cd) \neq (ab)(cd)(abc)$ .

Therefore $A_5$ doesn't have a normal subgroup of order $2$.

Is this proof correct?

Can someone please provide some alternative proof for this question .

  • $\begingroup$ Looks ok to me. Depending on how many times it has been covered (and whether it happened in this course or on an earlier course) a teacher might want you to justify the step It can be shown that .... It does sound like this poses no problem to you. $\endgroup$ – Jyrki Lahtonen Jul 2 '18 at 6:54
  • $\begingroup$ Well, that was easy :-) $\endgroup$ – Jyrki Lahtonen Jul 2 '18 at 6:55

A subgroup $H$ of $G$ is normal if and only if $H$ is the union of conjugacy classes of $G$ (For proof, see this link)

So computing normal subgroups in $S_n$ and $A_n$ are easy! Since elements of $S_n$ are conjugate if and only if they have the same cycle type.

  • $\begingroup$ But there are a few subtleties (not really relevant here) when figuring out conjugacy classes in $A_n$ as opposed to $S_n$. $\endgroup$ – Jyrki Lahtonen Jul 4 '18 at 9:50

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