TITLE: How many distinct subgroups does the symmetric group $S_3$ have?
QUESTION [19 upvotes]: How many distinct subgroups does the symmetric group $S_3$ have?
So when I do this question, is there a general technique to use?
I know that the order of a subgroup divides that of the group.
But the because $S_3$ is not cyclic, there may not be bijection between divisors of $6$ and
the subgroups of it. (Or there is?)
So is the only way by trial and error?
For $S_3$ it may help to know that it is equal to the group of symmetries of a regular triangle. 
But for $S_4$ and higher, is there any simpler method?
Anyway, is the answer $6$?

REPLY [8 votes]: An important theorem to use is that the order of any subgroup must divide the order of $S_3$. Now, the order of $S_3$ is just 3! = 6. Hence, any candidate subgroup must have order 6, 3, 2, or 1. Note that the group of order 1 is just the trivial subgroup { e }, the group containing the identity element of $S_3$. Now, this is the only subgroup of order 1. The subgroup of order 6 must be $S_3$ itself, the only possible subgroup of order 6. Hence, we are left with subgroups of order 2 or order 3.
Subgroups of order 2: Note any subgroup must contain the identity element e.
Hence, we are interested in sets of the form K = { e $\in $ K and |K| = 2 and s $\in$ S$_3$, such that s$\circ$s=e }. This is because we want closure and we need an identity and inverse for any subgroup. It is not difficult to check that the only elements in $S_3$ satisfying the conditions on K are the transpositions on $S_3$.
Subgroups of order 3: For subgroups of order three we are interested in sets of the form F = { e $\in$ F and |F| = 3, and $s_1$,$s_2$ $\in$ S$_3$ such that $s_1$$\circ$$s_2$=$s_2$$\circ$$s_1$= e }. This is because we want to make sure our set is closed, and contains inverses and the identity. It is not difficult to check that there are only two elements in $S_3$ with unique inverses (inverses that are not the element itself) that satisfy all of the conditions imposed on F.
So you are done, and the number of unique subgroups of $S_3$ is 6.