# Orders of the elements in the alternating group $A_5$

Find all possible orders of the elements in $$A_5$$

My idea for solving this was showing that:

• any $$k$$-cycle can be written as a product of $$k-1$$ jointed $$2$$-cyles, like $$(ij)(ik)=(ikj)$$, and since the $$2$$-cycles are all odd permutations, that excludes all $$k$$-cycles where $$k$$ is even;
• for permutations which are products of disjointed cycles (which is only possible for two $$2$$-cycles or a $$3$$-cycle and a $$2$$-cycle, of orders $$2$$ or $$6$$), show an example for each of why it is/isn't in $$A_5$$

So my answer would be $$\{ 1,2,3,5\}$$. The thing is the whole process seems very tedious if we're talking about larger groups.

For finding the orders of the elements in $$A_8$$, this implied checking all orders from $$1\to 15$$, which isn't that hard, but isn't there another way to do this?

Using generating functions, you can write

$$\prod_{k=1}^n\left(1+z_k\left(x^ky^{k-1}+\left(x^ky^{k-1}\right)^2+\cdots\right)\right)=\prod_{k=1}^n \left(1+\frac{z_k}{1-x^{-k}y^{1-k}}\right)$$

and read off the possible cycle combinations in $$A_n$$ from the terms with $$x^ny^j$$ with even $$j$$. You can set $$z_k=1$$ since cycles of length $$1$$ don’t affect the order. For instance, for $$n=5$$ this calculation yields

$$x^5\left(1+z_2y^2+z_3y^2+z_5y^4\right)\;,$$

in agreement with your result. For $$n=6$$ we get

$$x^6\left(1+z_2z_4y^4+z_2y^2+z_3y^2+z_3y^4+z_5y^4\right)\;,$$

so the set of orders in this case is $$\{1,2,3,4,5\}$$. For $$n=8$$, Wolfram|Alpha doesn’t play along, but we can do the easy part for the $$7$$-cycles (which can occur) and the $$8$$-cycles (which can’t occur) by hand and use the $$x^8$$ term in the above calculation for the rest:

$$x^8\left(1+(z_2+z_2z_3+z_2z_4)y^4+z_2z_6y^6+z_2y^2+z_3z_5y^6+z_3y^2+z_3y^4+z_4y^6+z_5y^4\right)\;.$$

Thus the set of orders in $$A_8$$ is $$\{1,2,3,4,5,6,7,15\}$$.

• This is interesting (+1). Where might I learn how to manipulate generating functions like this? – Shaun Jan 25 '20 at 12:29
• @Shaun: A good place to start might be generatingfunctionology. – joriki Jan 25 '20 at 12:30
• Excellent, thank you! $\ddot\smile$ – Shaun Jan 25 '20 at 12:31