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 $\{1,2,3,4,5,6,7,15\}$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.