Problem

Show that an element has order \(2\) in \(S_n\) if and only if its cycle decomposition is a product of commuting \(2\)-cycles.

Solution

The cycle decomposition of any permutation \(\sigma \in S_n\) is a product of disjoint cycles. Further, disjoint cycles commute. Therefore, “commuting” in the above statement is redundant. The statement we are proving is equivalent to:

\(\sigma \in S_n\) has order \(2\) if and only if its cycle decomposition is a product of \(2\)-cycles.

Let \(\sigma = \sigma_1\sigma_2…\sigma_n \in S_n\) be the cycle decompsition of \(\sigma\). Again, \(\sigma_i\)s are disjoint and they commute.

\(\Leftarrow\)

Suppose that \(\sigma\)’s cycle decomposition is a product of \(2\)-cycles. Then, every \(\sigma_i\) is a \(2\)-cycle.

Since each \(\sigma_i\) is a \(2\)-cycle, \(\sigma(x) \ne x\) for all \(x \in \{1, 2, …n\}\). Therefore, the order of \(\sigma\) is at least \(2\).

Since all \(\sigma_i\)s commute and are \(2\)-cycles, it follows that

$$ \sigma^2 = \sigma_1\sigma_2…\sigma_n\sigma_1\sigma_2…\sigma_n = \sigma_1^2\sigma_2^2…\sigma_n^2 = 1 $$

Thus, \(\sigma\) has order \(2\).

\(\Rightarrow\)

Conversely, suppose that \(\sigma\)’s cycle decomposition is not a product of \(2\)-cycles. That is, there exists \(i\) such that \(\sigma_i\) is an \(n\)-cycle for some \(n \ge 3\).

Then for any \(x\) appearing in \(\sigma_i\),

$$ \sigma^2(x) = \sigma_i^2(x) \ne x $$

Therefore, \(\sigma^2 \ne 1\) and thus the order of \(\sigma\) is not \(2\).

$$\tag*{$\blacksquare$}$$