How many elements of order $2$ does Sym $6$ have?

One can always count all the elements of order $2$ in $S_6$.

Since we have six elements to play with, it's clear that elements of order two look either like $(ab)$, $(ab)(cd)$ or $(ab)(cd)(ef)$.

  • First we find all the elements that look like $(ab)(cd)(ef)$. There are ${6\choose 2} = 15$ ways to create a 2-cycle $(ab)$. To get $(ab)(cd)$ we have ${4\choose 2}=6$ options left. Next, there's only one possible choice left to get $(ab)(cd)(ef)$. Finally, by canceling the $3!$ repetitions due to ordering 3 cycles, we get $$ \frac{15 \times 6 \times 1}{3!} = 15 $$

  • Similarly, for $(ab)(cd)$ we have $$ \frac{15 \times 6}{2!} = 45 $$

  • Finally, for $(ab)$ we have $$ {6\choose 2} = 15 $$

Therefore there are $15 + 45 + 15=75$ elements of order $2$ in $S_6$.


One way is to use GAP, like so:

gap> G:=SymmetricGroup(5);
Sym( [ 1 .. 5 ] )
gap> Ord2:=[];
[  ]
gap> for g in G do if Order(g)=2 then AddSet(Ord2, g); fi; od; Print(Size(Ord2));
25

and

gap> H:=SymmetricGroup(6);
Sym( [ 1 .. 6 ] )
gap> Order2:=[];
[  ]
gap> for h in H do if Order(h)=2 then AddSet(Order2, h); fi; od; Print(Size(Order2));
75

The sequence of these numbers is OEIS A001189; there apparently isn’t a really nice closed form, but there is a recurrence that is easy to use.

Let $a_n$ be the number of elements of order $2$ in $S_n$. Each such element $\pi\in S_n$ gives rise to a corresponding element of $S_{n+1}$ that agrees with $\pi$ on $[n]$ and fixes $n+1$. Each of the remaining elements of order $2$ in $S_{n+1}$ must be either a transposition $(k,n+1)$ for some $k\in[n]$, or the product of such a transposition a permutation of $[n]\setminus\{k\}$ of order $2$. There are $a_{n-1}$ permutations of $[n]\setminus\{k\}$ of order $2$, so

$$a_{n+1}=a_n+n(1+a_{n-1})\;,$$

where $a_1=0$ and $a_2=1$.

Tags:

Permutations

Group Theory

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