2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ In your application of Lagrange's theorem, you seem to have overlooked subgroups of order 3. Of course these are not simple either, but you must treat this case. $\endgroup$ – Doc Nov 19 '13 at 6:57
2
$\begingroup$

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 $A_4$

$\endgroup$
  • $\begingroup$ Hey, thanks for the input. I haven't learned about conjugacy classes yet, so I don't know what you're talking about unfortunately. $\endgroup$ – John Nov 2 '13 at 18:49
  • $\begingroup$ @John: I have expanded on my answer. A conjugacy class is a simple idea, and is intertwined with normality. Have a look. $\endgroup$ – Prahlad Vaidyanathan Nov 2 '13 at 19:00
  • $\begingroup$ Thank you for the detailed answer! $\endgroup$ – John Nov 2 '13 at 19:13
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.