Assumption in proof: The alternating group $A_n$ is simple.

Let $N$ be a nontrivial normal subgroup of $A_n$, $n \ge 5$, where $A_n$ denote the alternating group.

The book want to prove that $N$ contains a 3-cycle. (Niels Lauritzen, Concrete Abstract Algebra.)

Let $\sigma \in N, \sigma \ne e$. Then I can write $\sigma = \tau_1 \tau_2 \cdots \tau_r$ as a product of disjoint cycles.

Now comes the step:

Assume $\tau_1, \tau_2$ are both transpositions, then $\sigma = (a b)(c d) \cdots \tau_r$.

... Then we may assume that $\tau_1 = (a b) = (1 2)$ and $\tau_2 = (c d) = (3 4)$.

I know that there exists $\lambda \in S_n$ such that $\beta =\lambda \sigma \lambda^{-1}=(1 2)(3 4) \cdots \lambda \tau_r \lambda^{-1}$ is still a product of disjoint cycles.

Since $\sigma \in N$ this implies $\sigma \in A_n$ so $\beta \in A_n$.

How do I show that $\beta \in N$ or $\lambda \in A_n$ ?

Picture of proof:

/Nicolas

• While your question is a good algebraic question in and of itself, I don't think it's necessary to solve the problem. Informally, "$(12)(34)$ without loss of generality" is just trying to relabel the objects $S_n$ is acting on to avoid notational annoyances. Formally, $\lambda A_n \lambda^{-1} = A_n$ since $A_n$ is normal (index 2) in $S_n$; even if $N' := \lambda N \lambda^{-1} \ne N$, it still should be the case that $N'$ is a normal subgroup of $A_n$ that is isomorphic to $N$, and you can argue the properties of $N'$ and then infer properties of $N$. – Greg Martin Oct 28 '13 at 13:18
• Hi Greg. Thanks for your answer. I've attached a picture of the proof. Do you see where I've lost the point ? It's really bothering me that I can't find the proper element in $A_n$ to transform $\sigma$. – Shuzheng Oct 28 '13 at 13:34
• Greg has it right. You could say that the transpositions are $(ij)$ and $(kl)$, and then proceed to give the argument with $i,j,k,l$ in place of $1,2,3,4$, but it doesn't add any real substance to the argument, so it is convenient just to relabel. – Cheerful Parsnip Oct 28 '13 at 22:28
• Ahh nice ! Asked my teacher, but he said that I could always find a $\sigma \in S_n$ to transform the transpositions accordingly. However I guess he is not right about this. – Shuzheng Nov 1 '13 at 7:07
• @NicolasLykkeIversen Well, he is right based on the fact that $A_n\lhd S_n$, cf. Greg's comment. – Hagen von Eitzen Oct 26 '15 at 15:14