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 $6$?
Best Answer
I think that for small groups like $S_3$ this not very difficult to do by hand.
$S_3=\{id,(12),(13),(23),(123),(132)\}$
Clearly, $\{id\}$ and $S_3$ are subgroups of $S_3$
Now let us have a look at subgroups generated by one elements (cyclic subgroups). (We only have 6 elements to try, and we have already considered $[id]=\{id\}$.)
$[(12)]=\{id,(12)\}$
$[(13)]=\{id,(13)\}$
$[(23)]=\{id,(23)\}$
$[(123)]=[(132)]=\{id,(123),(132)\}$
Is this list of all subgroups? Yes, it is.
We can do this by trial and error -- trying adding one element to subgroup from our list and checking whether they generate the whole group $S_3$. (Which seems tedious, but in fact we only have to check whether any subgroup containing at least one 2-cycle and at least one 3-cycle is the whole $S_3$; and also observe that two different 2-cycles give us a 3-cycle.)
But quicker way is to notice that any proper subgroup of $S_3$ must have 1, 2 or 3 elements. (Order of subgroup must divide order of the group.) And any group of prime order is cyclic. So by enumerating cyclic subgroups we have obtain list of all proper subgroups of $S_3$.
We can check our result at groupprops wiki (or elsewhere): Subgroup structure of symmetric group:S3.
You also asked about $S_4$. As I have mentioned in my comment, there is another question on this site about subgroups of $S_4$: How to enumerate subgroups of each order of $S_4$ by hand