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Isaacs 6.13: If $\textbf{Z}(A_n)>1$, then $n=3$.

I am unsure where to begin. I know that $\textbf{Z}(A_n)$ is the center of $A_n$. If we assume that $\textbf{Z}(A_n)>1$, then the center is not the trivial center (i.e. it doesn't only contain the identity element.

But how do I proceed from there?

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You know (I hope) that the center is a normal subgroup of $G $ then if you have that your group is simple you get that it's center must be trivial, either $G$ ( if $G$is abelian) or the identity element $e $.

We know $A_n $ is not abelian so it's center is not the whole group, as $A_n $ is simple for $n>4$ (you may google this fact if you don't knew about it) for $n>4$ $Z (A_n)$ has only one element, the identity.

Now we must check the cases $n=1,2,3,4$.

For $n=1$ the center is equal to the identity because the whole group is the identity.

For $n=2$ it only contains the identity because the group has only the identity element.

For $n=3$ it's easy to see that there is more than one element in the center, because $A_3 $ has order $3$ so it is a p-group.

For $n=4$ you may check that has trivial center easily. (Suppose there is an element that conmutes with every other element and get a contradiction).

Then we conclude that we must have $n=3$.

Hope it helps.

Jonathaniui
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