# Klein 4 group is the only proper normal subgroup in $A_4$

How do you show that the only nontrivial normal subgroup of $A_4$, which is also not the whole group is the Klein 4 group, denoted by $V$ (or isomorphic to the Klein 4 group)?

I've shown before that $V$ is a normal subgroup of $S_4$, and that $V \subset A_4$. Is there a way to use those facts?

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If $A$ is normal in $B$ and is contained in $C$ which is a subgroup of $B$, then $A$ is normal in $C$. Can you prove this? – Gerry Myerson Dec 3 '12 at 4:22
Yes, that's pretty clear to me. But how would that force $V$ to be the only normal subgroup that's nontrivial and not the whole group? – chubbycantorset Dec 3 '12 at 4:25
It doesn't. It just shows $V$ is normal in $A_4$. Other methods are needed to show it's the only such. See the answers. – Gerry Myerson Dec 3 '12 at 4:33

1) A subgroup of a group is normal in it iff it is a union of conjugacy classes

2) Two permutations in $\,S_n\,$ are conjugate iff they have the same cycle decomposition (i.e., the same lengths of cycles and the same ammount of cycles of each length)

3) A conjugacy class of an even permutation in $\,S_n\,$ remains exactly the same class in $\,A_n\,$ unless all the disjoint cycles in the cyclic decomposition of the permutation are of odd and different lengths, in which case the equivalence class splits in two classes in $\,A_n\,$

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It's possible that the question has arisen before conjugacy classes have been discussed. – Gerry Myerson Dec 3 '12 at 4:32
Perhaps, but it is my personal experience, both as student and lecturer, that permutation groups usually appear after conjugacy is learnt. – DonAntonio Dec 3 '12 at 4:33
I'm not too familiar with applications of conjugacy classes, but would the Sylow theorems help here? We do know that the order of $A_4$ = 12 = 3*2^2, and the order of $V$ = 2^2. – chubbycantorset Dec 3 '12 at 4:35
Well, yes: if you do what Gerry suggested and write down all the subgroups, then you'll realize $\,V\,$ is the only Sylow 2-subgroup of $\,A_4\,$. Checking the other subgroups will be easy to realize that is the only non-trivial one. One more hint, which follows from this of course: $\,A_4\,$ has no subgroup of order $\,6\,$ – DonAntonio Dec 3 '12 at 4:42
Thanks, really helped to know that $A_4$ doesn't have a subgroup of order 6. – chubbycantorset Dec 3 '12 at 4:50

First, you find all the subgroups of $A_4$. It's not that hard --- there aren't all that many of them. Then you look at each in turn and work out whether or not it is normal. If you have any trouble along the way, write back.

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I actually haven't been taught how to find all subgroups of a given group. Is there a method for doing it, or do you just have to brute force it? – chubbycantorset Dec 3 '12 at 4:37
In this case it is worth to brutal force it. – Bombyx mori Dec 3 '12 at 4:42
Well, it's an educated brute force. Show that if a subgroup contains two $3$-cycles $a$ and $b$ with $a\ne b^2$ (for example, $a=(1\ 2\ 3),b=(1\ 2\ 4)$) then it's the whole group; show that if a subgroup contains a $3$-cycle and a product of transpositions (for example, $a=(1\ 2\ 3),b=(1\ 2)(3\ 4)$) then it's the whole group; since every group element is of the type $(1\ 2\ 3)$ or $(1\ 2)(3\ 4)$ (or is the identity), that doesn't leave much else to try, as far as subgroups go. – Gerry Myerson Dec 3 '12 at 9:50