Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to show that the Klein four-group is a normal subgroup of the alternating group $A_4$.

I am using the information in this link, that shows explicitly $A_4$, and Klein four-group as a subgroup.

I know that there is the direct way, by definition, but is there a way that does not require actually multiplying so many permutations ?

share|improve this question
    
@Chris Eagle - forgot it, I added it to the post –  Belgi Mar 17 '12 at 21:50
add comment

2 Answers

up vote 4 down vote accepted

The elements of the Klein $4$-group sitting inside $A_4$ are precisely the identity, and all elements of $A_4$ of the form $(ij)(k\ell)$ (the product of two disjoint transpositions).

Since conjugation in $S_n$ (and therefore in $A_n$) does not change the cycle structure, it follows that this subgroup is a union of conjugacy classes, and therefore is normal.

share|improve this answer
    
I will accept this after the timer allows me. I forgot that conjugation does not change the cycle structure.. Thank you! –  Belgi Mar 17 '12 at 21:55
1  
@Belgi: Also, remember that in general it suffices to show that $ghg^{-1}\in H$ and $g^{-1}hg\in H$ for every $h$ in a generating set for $H$ and every $g$ in a generating set of $G$ in order to show that $H$ is normal in $G$; so you often don't have to compute all possible $ghg^{-1}$, just some. –  Arturo Magidin Mar 17 '12 at 22:25
    
thank you again. Is there a simple way to see why does this subgroup is closed under multiplication ? –  Belgi Mar 17 '12 at 22:44
    
@Belgi: Which group? The Klein $4$-group inside $A_4$? Noting that $(12)(34)\circ(13)(24) = (14)(23)$ is sufficient, together with the fact that each of them is its own inverse. –  Arturo Magidin Mar 17 '12 at 22:46
    
Pretty solution! –  Bombyx mori Jul 1 '13 at 3:28
add comment

You do not have to compute the entire Cayley table of a group to get an isomorphism. I think that this is what you are struggling with so I'll put a little of detail.

First of all, there are only two groups of order $4$ : the cyclic group of order $4$ and the Klein group, it is quite easy to see it. Let $G$ be a group of order $4$ : by Cauchy's theorem, since $2$ divides $4$ and $2$ is prime, there exists an element of order $2$ in $G$. Let $a$ be this element and $H = \langle a \rangle$. Since $[G:H] = 2$, $H \triangleleft G$, hence for all $g \in G$, $gag^{-1} \in H$, but since $a \neq 1$, we must have $gag^{-1} = a$, that is, $a$ commutes with every element of $G$.

Consider an element outside $H$, call it $b$. If it has order $4$, then $G = \langle b \rangle \cong C_4$, the cyclic group of order $4$. If not, then since the order of an element divides the order of the group, and $b \neq 1$, then $b$ must have order $2$. But then the fourth element cannot have order neither $1$ or $4$, hence it must have order $2$ too. Hence every non-trivial element of $G$ has order two, and using the argument above, they commute with each other ; this gives you the group $C_2 \times C_2$, which is precisely the Klein group, because we must have $ab = c$, $ac = b$ and $bc = a$ (for obvious reasons, because other possibilities lead to contradictions).

Now the subgroup of $A_4$, namely $K=\{(1), (12)(34),(13)(24),(14)(23) \}$ is a subgroup of order $4$. Since it is not cyclic, it is isomorphic to the Klein group. Conjugation in $S_n$ does not change cycle structure, so that in particular it does not do that in $A_n$. This means that this subgroup is normal, because $gKg^{-1} \subseteq K$, which is an equivalent condition for normality of a subgroup.

Hope that helps,

P.S. : Maybe I used sledgehammers to classify groups of order $4$, but I just threw out the first ideas that came to mind ; if you want to simplify them by commenting feel free.

share|improve this answer
    
Why would you want to classify groups of order 4? I don't think it is relevant. Furthermore, the end of the argument is just a rewrite of Arturo's answer... –  M Turgeon Mar 17 '12 at 22:06
    
Because then looking at this subgroup $K$ of $A_4$ you know it is isomorphic to $C_2 \times C_2$ by noticing that there is no element of order $4$ in this subgroup ; instead of computing the isomorphism by hand (which OP asked for, i.e. "is there a way without multiplying so many permutations"), I only use the fact that I know which are the groups of order $4$. The end of the argument is just a re-write indeed, but it is not the point of my answer. –  Patrick Da Silva Mar 17 '12 at 22:18
    
The question was not "what is the isomorphism type of this subgroup", but rather "how can I check it is normal without computing 'so many' products?" (i.e., without having to check $ghg^{-1}\in H$ for every $h\in H$ and possibly every $g\in G$). –  Arturo Magidin Mar 17 '12 at 22:24
    
I didn't understand the question well enough without the link, because I didn't know that OP knew that the Klein group was actually isomorphic to the subgroup, which was the question I answered. Now that OP added his link I took a look and obviously I answered a different question than his. –  Patrick Da Silva Mar 17 '12 at 22:29
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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