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The exponent of a group $G$, denoted $\text{Exp}(G)$, is the smallest $n\in \mathbb{N}$ such that $x^n=1$ for every $x\in G$.

This page says that $$\text{Exp}(A_n)=\left\{\begin{array}{lcl}\text{lcm}\{1,\ldots,n-1\}&:&n \text{ is even}\\\text{lcm}\{1,\ldots,n\}&:&n \text{ is odd} \end{array}\right.$$

But I think this must be false. $A_3$ is just the cyclic group $C_3$ so its exponent is $3$, but $\text{lcm}(1,2,3)=6$. Then $A_5$ has exponent $30$, but $\text{lcm}(1,2,3,4,5)=60$. It seems to be right for even $n$, so one would think perhaps it's just off by a factor of $1/2$ for the odd $n$ - but then, it works for $A_7$! After that it goes back to being off again with $\text{Exp}(A_9)=1260$ whereas $\text{lcm}\{1,\ldots,9\}=2520$.

So, what is the formula for $\text{Exp}(A_n)$? How can we prove it?

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What you say "this page" says is wrong: for odd $\,n\,$ that page says the exponent is $\,lcm\{1,2,...,(n-2),n\}\,$ , which for $\,n=3\,$ gives $\,3\,$, precisely. – DonAntonio Feb 2 '13 at 10:59
@DonAntonio Actually this is because I sent the correction to the site owner (with the proof that Mariano helped me figure out). – Samuel Handwich Feb 3 '13 at 18:21
up vote 2 down vote accepted

The order of a permutation is the lcm of the sizes of the cycles appearing in its disjoint cycle factorization. The lcm of the alternating group is, then, the lcm of the sizes of the cycles of the disjoint cycle decompositions of its elements.

If $k\leq n$ is such that either $k$ is odd or $k+2\leq n$ then there is a permutation in $A_n$ which has a cycle of length $k$; indeed, in the first case, we can take it to be a cycle, in the second case a cycle times a transposition. On the other hand, if $k\leq n$ is even and $k+2>n$, then we must have either $k=n-1$ and $n$ odd or $k=n$ and $n$ even, and there are no elements in $A_n$ with a cycle of size $k$.

  • Show that this gives the sizes of all possible cycles appearing in elements of $A_n$.

  • Find a formula for the number you want.

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"then there is a permutation in $A_n$ which has a cycle of length $k$" You mean that there is a $k$-cycle in the permutations decomposition into disjoint cycles? – Samuel Handwich Feb 2 '13 at 4:07
Indeed. ${}{}{}$ – Mariano Suárez-Alvarez Feb 2 '13 at 4:07
Sorry I am still not understanding this. What formula am I supposed to be getting here? I just keep coming up with $\text{lcm}\{1,\ldots, n-2\}$ which is obviously not right. – Samuel Handwich Feb 2 '13 at 4:26
As I said, find the size of the cycles appearing in elements of $A_n$ (for example, using the observations I made in the 2nd paragrah) The exponent of $A_n$ is then the LCM of all those numbers. – Mariano Suárez-Alvarez Feb 2 '13 at 4:27
Ahhhh okay, I got it: $\text{lcm}\{1,\ldots,n-2,n\}$ for odd $n$. Thank you. – Samuel Handwich Feb 2 '13 at 4:36

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