1
$\begingroup$

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?

  • 0
    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.2012-12-03

2 Answers 2

3

The following may help you:

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 $\,A_n\,$

  • 1
    @1ENİGMA1 Among other reasons, because conjugate elements in a group have the same **order**, and the order of $\;(123)\;$ is three, whereas the order of all the non-unit elements in the Klein's group is \two...2018-01-09
1

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.

  • 0
    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.2012-12-03