Let \(G\le S_6\) and suppose \(G\) contains an element of order \(6\). Prove that \(G\) has a normal subgroup of index \(2\).
Solution
1. Every permutation of order \(6\) in \(S_6\) is odd.
Let \(\sigma\in S_6\) have order \(6\). The order of a permutation is the least common multiple of the lengths of the cycles in its disjoint-cycle decomposition.
Since \(3\) divides the order of \(\sigma\), its cycle decomposition contains a cycle of length divisible by \(3\). With only six letters, such a cycle has length \(3\) or \(6\).
If \(\sigma\) contains a \(6\)-cycle, then it is itself a \(6\)-cycle. Its sign is \[ (-1)^{6-1}=-1. \]
Otherwise \(\sigma\) contains a \(3\)-cycle. Since its order is also divisible by \(2\), the remaining three letters must contain an even cycle. The only possibility is a \(2\)-cycle, leaving one fixed point. Thus the cycle type is \[ 3+2+1. \] A \(3\)-cycle is even and a \(2\)-cycle is odd, so their disjoint product is odd. Hence every order-\(6\) element of \(S_6\) is odd.
2. The restriction\[ \operatorname{sgn}|_G:G\longrightarrow\{\pm1\} \]is surjective.
By hypothesis, \(G\) contains an element \(\sigma\) of order \(6\). Step 1 gives \[ \operatorname{sgn}(\sigma)=-1. \] The identity element maps to \(1\), so the image contains both elements of \(\{\pm1\}\).
3. The subgroup\[ H\coloneqq G\cap A_6 \]is normal in \(G\) and has index \(2\).
Since \[ H = \ker(\operatorname{sgn}|_G), \] it is normal in \(G\). By step 2, the restricted sign homomorphism has image of order \(2\). The first isomorphism theorem gives \[ G/H\cong\{\pm1\}, \] so \[ [G:H]=2. \]
4. Q.E.D.
Step 3 supplies the required normal subgroup of index \(2\).