# Thread: Abelian subgroups of S_{n}

1. ## Abelian subgroups of S_{n}

I am finding all abelian subgroups of a symmetric group $S_{n}$ for n>=5.

What I have found so far is
1. {e} : trivial group
2. a cyclic group of order n
3. a quotient group of order 2: $S_{n}/A_{n}$.

I am wondering if below 4 is correct.
4. "a cyclic group of order k which is a divisor of n"

Any more subgroup exists for $S_{n}$?

2. Originally Posted by aliceinwonderland

I am finding all abelian subgroups of a symmetric group $S_{n}$ for n>=5.
i don't think this is basically possible to do because we even don't know all cyclic subgroups of $S_n.$ no formula is known for the maximum order of elemens of $S_n.$ however, there's a result due

to Landau that if $f(n)$ is the maximum order of elements of $S_n,$ then $\lim_{n\to\infty} \frac{\ln f(n)}{\sqrt{n \ln n}} = 1.$ it's not surprising that there's no closed form for $f(n),$ because $f(n)$ is related to the partition function.

3. Originally Posted by NonCommAlg
i don't think this is basically possible to do because we even don't know all cyclic subgroups of $S_n.$ no formula is known for the maximum order of elemens of $S_n.$
Any concrete example of an element of $S_n$ that has an order bigger than n? It is easy to find an element of order n in $S_n$, but it is hard for me to find an element that has an order bigger than n.

4. Originally Posted by aliceinwonderland
Any concrete example of an element of $S_n$ that has an order bigger than n? It is easy to find an element of order n in $S_n$, but it is hard for me to find an element that has an order bigger than n.
Sure, take $S_7$ and consider $(1234)(567)$.