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.

Let $N$ a non-trivial normal subgroup of $A_n$ and $H = N \cap A_{n-1}$. I would like to show that $A_{n-1} \hookrightarrow A_n \to A_n/N$ is surjective, where $A_n \to A_n/N$ is the canonical homomorphism, and that the kernel of this composite map is precisely $H$. Any help will be greatly appreciated!

share|improve this question
    
For $n>4$, $A_n$ has no non-trivial normal subgroups (other than itself), so then $A_n/N = A_n/A_n$. It would be trivially true for the second step mapping. So you just need to check cases for $n<5$. –  gnometorule Dec 3 '12 at 4:44
    
Yes, but he is not allowed to use the simplicity of $A_n$. Perhaps that is what he is trying to prove? –  Derek Holt Dec 3 '12 at 9:54
add comment

2 Answers

Directly using the isomorphim theorems. The image of $\,A_{n-1}\,$ in $\,A_n/N\,$ is

$$A_{n-1}N/N\cong A_{n-1}/\left(A_{n-1}\cap N\right)=A_{n-1}/H$$

share|improve this answer
    
How does this imply that the map is surjective? How does this argument use any of the specifics of the problem -- the statement wouldn't be true for just any subgroup instead of $A_{n-1}$? –  joriki Dec 3 '12 at 4:51
    
Well, I didn't read that carefully the quetion, true, but that map is surjective on the above, not on $\,A_n/N\,$ –  DonAntonio Dec 3 '12 at 4:53
    
Why not on $A_n/N$? –  joriki Dec 3 '12 at 5:34
    
Well, at least a priori and without assuming we already know $\,A_n\,$ is simple for $\,n\geq 5\,$: if there's an actual, non-onto, embedding $\,A_{n-1}\hookrightarrow A_n\,$, and $\,N\triangleleft A_n\,$ , then the canonical projection $\,A_{n-1}\to A_{n-1}N/N\,$ is not onto a priori, and in the only non-trivial case, with $\,n=4\,$, we get $\,N\cap A_3=1\Longrightarrow A_4/N\cong A_4\,$ and thus we have a non-onto composition $\,A_3\hookrightarrow A_4\to A_4/N=A_4\,$ ... –  DonAntonio Dec 3 '12 at 11:53
add comment

The kernel of the composite map is precisely $H$ simply because $H\subseteq N$ so $H$ is mapped to $N$, and any element of $A_{n-1}$ that isn't in $H$ isn't in $N$ and hence isn't mapped to $N$.

For surjectivity, we need to show that every coset of $N$ contains at least one element that leaves $n$ fixed, that is, for any $g\in A_n$ there is $h\in N$ such that $h(g(n))=n$. If $g(n)=n$, then $h=e$ works; otherwise, since $N$ is non-trivial it contains at least one permutation that maps at least one element to a different element, and since $N$ is normal we can renumber this permutation such that these elements are $g(n)$ and $n$, respectively. The renumbering can be performed with an even permutation as long as $n\ge4$, so the statement needs to be checked separately for $n\lt4$.

share|improve this answer
1  
Surjectivity follows easily from the fact that $N$ is transitive. Stefan was asking how to prove that $N$ is transitive in an earlier post, so he is clearly meant to be using that fact here. –  Derek Holt Dec 3 '12 at 9:53
add comment

Your Answer

 
discard

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