Prove that a permutation in $S_n$ can be written as the product of disjoint transpositions if and only if it is of order $1$ or $2$.
What I did so far:
$\leftarrow$ if $s \in S_n$ is of order $1$ then that means $s=I$ and therefore it can be written as $(11)(22)…(nn)$, which are disjoint transpositions. If it is of order $2$ then it is made up of $1$-cycles and $2$-cycles (since the $lcm$ is the order).
And there is where I had problems. If it is of order $2$ then how can I write it as disjoint transpositions? Also, I am having a hard-time with the other direction.
Best Answer
Facts:
Every permutation can be written (essentially uniquely) as a product of disjoint cycles (of different lengths).
The order of a permutation is the lcm of the lengths of those cycles in this representation.
If $\sigma$ is a permutation of order 1 or 2, then $\sigma = c_1 c_2 \ldots c_k$, as a disjoint product of cycles. If one of the lengths of $c_i$ is $>2$, so is the lcm of their orders (it's a multiple of each of those lengths), and so the order of $\sigma$ would be $>2$, which cannot be. So all lengths of cycles $c_i$ are 1 or 2. If all are we have the identity, otherwise a product of disjoint transpositions (we have at least one of order 2). We can leave out the 1-cycles, as is customary.
This is basically the whole argument. The other way round, if $\sigma$ is a product of involutions (at least one non-trivial), the order is clearly 2 (as the lcm of 1's and 2's is just 2).