how many subgroups of order 3 are there in symmetric group $S_6$ Assuming $S_6$ is a symmetric group, determine how many subgroups of order $3$ are there in $S_6$?
What are the subgroups of order $3$ like? Is it like $\{(12)(34)(56)\}$?
 A: A subgroup of $S_6$ must, of course, contain the identity permutation id. 
A subgroup of order three, say $A\leq S_6$, must must then contain two additional permutations, $\varphi$ and $\tau$ such that $\varphi^3 = \tau^3 = id$, each of which generates $A$, and such that $\varphi$ and $\tau$ are inverses of each other. Take, for example, the two 3-cycles $\varphi = (1 , 2, 3)\in S_6$ and $\tau = \varphi^2 = (1, 3, 2) \in S_6$. They are inverses of each other; you can check $\varphi ^3 = \tau ^3 = id.$ So each is of order $3$, and $\varphi \circ \tau = (1, 2, 3)(1, 3, 2) = \tau \circ \varphi = id $, so they are inverses of each other. Hence $A = \{id, (1, 2, 3), (1, 3, 2)\}$ is a subgroup of $S_6$. 
Hence, $A$ is a subgroup of $S_6$ of order $3$. Your task is to determine how many such subgroups of order $3$ exist in $S_6$.
Theorem: The order of a cycle is equal to its length. The order of a product of disjoint cycles of lengths $i, j$ is equal to the $\text{lcm}(i, j)$. 
So in $S_6$, (a) all permutations that are cycles of length $3$, and (b) all permutations that are the product of two disjoint cycles each of length $3$, like, say, $(2, 4, 6)(1, 3, 5)$, are of order $3$, and no other cycle-types are. Every subgroup of order three contains two distinct 3-cycles and the identity, or else two distinct products of disjoint 3-cycles. 
Count all such permutations of type (a) or (b) in $S_6$  divide by 2, and you'll have the number of subgroups of $S_6$ whose order is $3$.
A: A subgroup of order $3$ is generated by an element of order $3$. Can you show that the only elements of order $3$ are $3$-cycles like $(123)$ and double $3$-cycles like $(123)(456)$? How many of these elements are there? The number of subgroups of order $3$ will be exactly half this number. (Can you see why?)
