# Finding element of maximal order in symmetric group

• Dec 13th 2009, 09:26 PM
jackie
Finding element of maximal order in symmetric group
Would someone show me the method to find an element of maximal order in $\displaystyle S_7$? Is there a general technique for this?
I think that this element could have order 12=3 $\displaystyle \times$4 since I can find a (4,3) cycle element in $\displaystyle S_7$ like (1234)(567) which has order 12.
• Dec 13th 2009, 10:20 PM
NonCommAlg
Quote:

Originally Posted by jackie
Would someone show me the method to find an element of maximal order in $\displaystyle S_7$? Is there a general technique for this?
I think that this element could have order 12=3 $\displaystyle \times$4 since I can find a (4,3) cycle element in $\displaystyle S_7$ like (1234)(567) which has order 12.

$\displaystyle 12$ is the answer. if there exists an integer $\displaystyle m=p_1^{k_1}p_2^{k_2} \cdots p_r^{k_r},$ where $\displaystyle p_1, \cdots, p_r$ are distinct primes, such that $\displaystyle p_1^{k_1} + p_2^{k_2} + \cdots + p_r^{k_r} = n,$ then the maximal order of elements in $\displaystyle S_n$ is $\displaystyle m.$

for example, since $\displaystyle 12=2^2 \times 3$ and $\displaystyle 2^2 + 3 = 7,$ the maximal order of elements in $\displaystyle S_7$ is $\displaystyle 12.$

see section 4 (the prime connection) of this paper for more details. don't get scared, it's very easy to understand! (Nod)