Showing There Is Only One Non-Trivial Normal Subgroup.

Question

Show that $H$ $=$ {$(id), (12)(34), (13)(24), (14)(23)$} is the only non-trivial normal subgroup for $A_4$.

Looking at this question, what I started with was:

By Lagrange's theorem, there are only subgroups of order: 1, 2, 4, 6 and 12. Clearly the trivial normal subgroups are $A_4$ and ${(id)}$ thus we can eliminate the subgroups of order 1 and 12.

From what I proved from an assignment a month ago, there doesn't exist a subgroup of order 6 for $A_4$ so we can eliminate that as well.

Now we have to deal with the subgroups of order 2 and order 4. I'm having a block here now. I know the condition of a normal subgroup, but I can't really think of a way to show that H is the only non-trivial normal subgroup.

Any hints? :O

Answer 1

Ok, here is a more detailed proof then :

If $N \triangleleft A_4$, then for any $x\in A_4$, consider the set $$ C(x):= \{gxg^{-1} : g\in A_4\} $$ (called the conjugacy class of $x$) By normality of $N, C(x) \subset N$.

Now, $|A_4| = 12$, so the possibilities for the orders of subgroups of $A_4$ are $\{1,2,3,4,6,12\}$. You have already eliminated $\{1,6,12\}$, leaving $\{2,3,4\}$.

Now, if $x \in A_4$ such that $|x| = 2$, then $x$ must be one of $$ (12)(34), (13)(24), \text{ or } (14)(23) $$ For any such $x$, $C(x)$ contains all the others (see this link for the details). Hence, $|N| \neq 2$.

If $x \in A_4$ such that $|x| = 3$, then $x$ must be in one of $$ \{(123), (132)\}, \{(124), (142)\}, \{(134), (143)\} $$ Again, you can check that if $x$ is any such element, its conjugacy class must contain an element from another of those sets. In particular, if $x\in N$, then there must exists $y\in N\setminus\{x,x^2,e\}$. Hence, $|N| \neq 3$.

This leaves only 4, and you can check that the subgroup $H$ you have described is normal in $A_4$

Answer 2

Only subgroups of order $2$ are $\{e, (12)(34)\},\{e, (13)(24)\},\{e, (14)(32)\}$

You can clearly see that none of these are normal..

If they are normal then their generators should be in the center.. But $Z(S_4)=\{e\}$

Now, only group of order $4$ is $\{(e), (12)(34), (13)(24), (14)(23)\}$ and so you are done.