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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.

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  • $\begingroup$ Do this for the symmetric group first. Then it will be clear. $\endgroup$ – Pedro Tamaroff Aug 1 '16 at 22:06
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Suppose that $n>2$ and pick any $\tau\in S_n$ that is not the identity. Then there is a pair $i<j$ such that $\tau (i)=j$. I claim there is $\sigma\in S_n$ that does not commute with $\tau$. Because $n>2$ there is $k$ distinct from $i$ and $j$. Let $\sigma=(jk)$. Then $\tau \sigma(i)=j$ while $\sigma\tau(i)=k$. This proves $S_n$ has trivial center for $n>2$.

Now pick any $\sigma\in A_n$ where $n>3$. We already know there is a transposition $(jk)$ that does not commute with $\sigma$, and because $n>3$ there are $m,l$ different from $j,k$ and from each other, so that $(ml)(jk)$ is even and does not commute with $\sigma$.

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    $\begingroup$ Consider $\sigma=(mj)(kl)$. Then $(jk)$ does not commute with $\sigma$, but $(ml)(jk)$ does, so your final claim is incorrect. $\endgroup$ – Akiva Weinberger Apr 2 '18 at 3:52
  • $\begingroup$ @AkivaWeinberger True. You need a somewhat hands on argument for $n=4$, but for $n>4$ you do have wiggle room. :) $\endgroup$ – Pedro Tamaroff Apr 2 '18 at 12:55
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    $\begingroup$ I wrote an answer below that seems simpler and doesn't have that problem $\endgroup$ – Akiva Weinberger Apr 2 '18 at 14:00
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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.

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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.

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    $\begingroup$ Well, the problem is that what I need to prove about center is used to prove "$A_n$ is simple for $n \geq 5$" theorem in my book. $\endgroup$ – Jxt921 Aug 1 '16 at 22:13
  • $\begingroup$ Oh, okay, that's too bad. Sorry. $\endgroup$ – GiantTortoise1729 Aug 1 '16 at 22:17

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