The center of $A_n$ is trivial for $n \geq 4$

Question

I need to prove that the center of $A_n$ is trivial for $n \geq 4$.

$Z(A_3) = A_3$, since $A_3 = \mathbb{Z}/3\mathbb{Z}$ is commutative.

One idea is two use "counting" technique. First of, all we count cojugacy classes of even permutations in $S_n$. It's the classes correspodning to the types $[\lambda_1, ..., \lambda_r]$ where $n \equiv r \mod 2$

Now, the conjugacy class in $S_n$ correspoding to the type $[\lambda_1, ..., \lambda_r]$ splits into two equal-sized classes in $A_n$ if $\lambda_1, ..., \lambda_r$ are distinct odd numbers. If they aren't then it is preserved in $A_n$.

Then we can use "the counting formula"( counting the number of elements in a conjugacy class of $S_n$ ) of elements in each "even" class of $S_n$ and divide one by $2$ if needed.

Still, I'm not sure if it can be done. Seems to be a lot of work there. Maybe there are other, he easier ways? Or, maybe, it is the way, in this case, I wouild appreciate any advices.

Answer 1

I will show that, for every $\sigma$ other than the identity, there is something in $A_n$ that does not commute with it.

Since $\sigma$ is not the identity, $\sigma$ maps some element $a$ into $b$ with $a\ne b$. Choose $c$ and $d$ not equal to $a$ and $b$ (which I can do because $n>3$). Then I claim $(bcd)$ does not commute with $\sigma$. Proof: $\sigma(bcd)$ maps $a$ into $b$, but $(bcd)\sigma$ maps $a$ into $c$.

Therefore, no $\sigma$ other than the identity commutes with every element of $A_n$. In other words, no $\sigma$ other than the identity is in the center of $A_n$. Thus, the only element of the center of $A_n$ is the identity. QED.

Answer 2

Recall the center is a normal subgroup. For $n \geq 5$, $A_n$ is simple. For $n=4$ you just have to find an element that does not commute with all of the Klein $4$ group.