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everyone! Would anyone be willing to give me any sort of help with the following question?

Let $n\ge 4$ and $A_n$ the alternating group. Let $N$ a non-trivial normal subgroup of $A_n$. Prove that the action of $N$ on $\{1,2,...,n\}$ is transitive.

Let me stress that one is NOT allowed to use the fact that $A_n, n\ge 5$ is simple.

Any help will be greatly appreciated!

MORE INFORMATION: Would showing that if $X$ is an $N-orbit$ then $gX$ is an $N$-orbit, where $g \in A_n$ and using Cardinality of a subset acted upon by the Alternating Group, $A_n$ help?

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It is sometimes easier to deduce what you are trying to prove from a more general result. The point here is that the group $A_n$ is 2-transitive for $n \ge 4$, and the property that all normal subgroups are transitive is true for 2-transitive groups. (In fact it is true for the more general class of primitive permutation groups, but you need not worry about that!) John's solution below is almost correct, but you need to conjugate $g$ by an element of $A_n$ that fixes 1 and maps 2 to $x$, and such a $g$ exists by 2-transitivity of $A_n$. –  Derek Holt Dec 2 '12 at 11:46
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2 Answers

up vote 1 down vote accepted

Take a nontrivial element $g\in N$. Suppose $g$ sends 1 to 2. For any $x\in\{3,\ldots,n\}$, note that $(2\ x)g(2\ x)^{-1}$ is an element of $N$ that sends 1 to $x$. So the orbit of 1 is $\{1,\ldots,n\}$, hence $N$ acts transitively.

Edit: Pardon my stupidity. Assuming $x\ne 3,4$, you could consider $((2\ x)(3\ 4))g((2\ x)(3\ 4))^{-1}$ instead. That'll leave out the case $n=4$ though.

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Nop. The element $\,(2x)g(2x)^{-1}\,$ cannot be in $\,N\,$ as it is not an even permutation. –  DonAntonio Dec 2 '12 at 11:32
    
Oops.. Come to think of it, I vaguely remember seeing a theorem along the lines of: two elements are conjugate in $A_n$ iff they are conjugate in $S_n$. Is this correct? –  john Dec 2 '12 at 12:17
    
No @john: the conjugacy class of an element in $\,S_n\,$ "passes as it is" to $\,A_n\,$ (when dealing with an even permutation, of course) iff the cycle decomposition of that element is not a product of disjoint cycles of different _odd_ length (a fixed point is considered an odd cycle, of course, which means that in the above are to be considered, among others, even permutations with only one single fixed point), otherwise the conjugacy class splits in two within $\,A_n\,$ –  DonAntonio Dec 2 '12 at 12:40
    
There is no problem with $n=4$: take $x,y\in\{3,\dots,n\}$, $x\neq y$ and look at $(y\ x\ 2)^{-1}g(y\ x\ 2)$. This will send $1$ to $x$. (Why don't edit your answer in order to have a clear and full answer to this question?) –  user26857 Dec 3 '12 at 15:35
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Well, since $\,A_n\,\,\,\,\,n\geq 5\,$ , is simple, the only example there is for your question is $\,A_4\,$ , and the only normal non trivial subgroup of this one is

$$\{(1)\,,\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\}$$

which is easily seen to be transitive on $\,\{1,2,3,4\}\,$

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I am very sorry about the confusion. We are not allowed to assume that $A_n, n\ge 5$ is simple at this point. –  user44069 Dec 2 '12 at 3:59
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