Element of maximal order in alternating group $A_{10}$: how to prove this is it?

I'm trying to solve this exercise here:

What's the maximum order of any element in $A_{10}$?

Here are my thoughts:

One way of doing this seems to be by writing down all possible elements in $S_{10}$ (I mean just per order, not actually every element) and then looking at those that are in $A_{10}$.

This seems a lot of work but for completeness I will post this here. So, (some) possible orders for elements in $S_{10}$ are:

$$ \begin{matrix} \text{ disjoint cycles } & \text{ order} \\ (1)(1)(1)(1)(1)(1)(1)(1)(1)(1) & 1\\ (1)(1)(1)(1)(1)(1)(1)(1)(2) & 2\\ (1)(1)(1)(1)(1)(1)(1)(3) & 3\\ (1)(1)(1)(1)(1)(1)(4) & 4\\ (1)(1)(1)(1)(1)(5) & 5\\ \dots & \\ (10) & 10 \\ (7)(2)(1) & 14\\ (5)(3)(2) & 30\\ (5)(3)(1)(1) & 15 \\ \end{matrix} $$

where $(k)$ denotes a cycle of length $k$.

So these are not all orders but I think I got the interesting ones. For example, I think $30$ is the highest possible order in $S_{10}$.

Also, I think $15$ is the highest order in $A_{10}$.

But my problem is I don't know how to prove it.

How can I prove that $15$ is the highest order in $A_{10}$?

And my other question is:

Is there a failsafe way to find elements of highest order? By that I mean one that makes sure I'm not missing any elements?

It seems kind of tricky: these are such large groups but they don't seem to contain any elements of large orders. Or do they? Maybe I'm missing something?

Solution 1:

An element of $S_n$ of maximal order can always be chosen to have a cycle structure in which all the cycle lengths are prime powers for distinct primes. For $A_n$ there may possibly be an additional $2$-cycle to make the permutation even. This appears if and only if another cycle with length a power of $2$ is used. (Since cycles of odd length are even permutations and cycles of even length are odd permutations.)

We'll begin with the problem for $S_n$ and then worry about $A_n$.

The order of a permutation is the lcm of its cycle lengths. If you have a permutation that is not of the specified form, you can always bring it to that form, without lowering its order, through the following steps. If powers of the same prime $p$ appear in the lengths of two different cycles, you can obtain a shorter permutation (i.e. one that moves fewer elements) by dividing the length of the cycle with the lower power of $p$ by that power. Also, if you have a cycle of length $ab$, where $a, b > 1$ are relatively prime, you can replace them with cycles of length $a$ and $b$ and get a shorter permutation.

Thus the problem for $S_n$ is to choose prime powers whose sum is $\leq n$ and whose product is as large as possible.

For $A_n$, it's almost the same, except that after you've performed the reductions above you may not have an even permutation. But any reductions that may have been performed will have freed up at least two elements that can then be used to make a $2$-cycle which will make the permutation even. The only apparent exception is if the only reduction to be performed is for a $6$-cycle ($a, b$ are $2, 3$), but in this case replacing a $6$-cycle with a $2$-cycle and a $3$-cycle doesn't actually change the permutation's parity.

So an element of $A_n$ of maximal order can always be chosen with relatively prime, prime-power cycle lengths and at most one additional $2$-cycle. In this case you either want to find the highest product possible of odd prime powers with sum $\leq n$, or the highest product possible of prime powers including a power of $2$ whose sum is $\leq n - 2$.

In the case of $A_{10}$, the possible prime powers are $2, 3, 4, 5, 7, 8, 9$. Then the possibilities are: