Find the number of elements of order two in the symmetric group $S_4$ of
all permutations of the four symbols {$1,2,3,4$}.
the order two elements are two cycles.number of $2$ cycles are $6$.but the given answer is $9$.where I am wrong?can anybody help me.
more generally is there any formula for the problem "Find the number of elements of order $r$ in the symmetric group $S_n$"
Best Answer
Hint: you're forgetting to include and count those permutations that are the product of two disjoint two-cycles in $S_4$:
$$(1\, 2)(3 \,4),\; (1\, 3)(2 \, 4),\; (1\, 4)(2 \,3)\; \in S_4,$$ which comprise three additional elements in $S_4$, each of order $2$, as are the six 2-cycles of order $2$ you counted.
Recall that the order of a permutation which is the product of disjoint cycles is equal to the $\;\operatorname{lcm}\;$ (i.e., the least common multiple) of the orders of its cycles.