[Artus]

Let S_3 be the symmetric group of degree 3. Final all the subgroups of S_3.

Answer:


Let g=

and let f=

Then S_3 = {identity, f, g, f^2, fg, gf}

I first found the cyclic subgroups:

(identity) = {identity}
(f) = (f^2) = {identity, f, f^2}
(g) = {identity, g}
(fg) = {identity, fg}
(gf) = {identity, gf}

My professor told me how to find the rest of the subgroups...he said that these (the cyclic subgroups and S_3 itself) were actually the only subgroups of S_3 and showed me how to prove that there weren't any other. He said it was alot easier than trying to randomly check if there were any other subgroups. I thought his proof made sense to me at first, but then I got confused again...so can anybody try to clarify it for me. This was his answer:

In order for form a new subgroup, we need to take one element from (f) and one element from the rest of the cyclic groups. Then (of course) we will still need to check those possibilities. After checking all the possibilities, we will know that there are no other subgroups. What confuses me is...why do we have to choose one from (f) and one from the rest? For instance, why don't we choose one from (g) and one from the rest?

[DPLSK]

(Original post by Artus)
Let be the symmetric group of degree 3. Find all all the subgroups of .

I would prove the assertion that is a proper subgroup of if and only if is cyclic. I would do this using the subgroup test.

Firstly, show that any cyclic subset of is a subgroup.

Secondly, we use proof by contradiction. Assuming the group is not cyclic then we argue that it must be .

Note: The contradiction here comes from the fact that we must have a 'proper' subgroup of .

I hope that helps.

Darren

PS: Alternatively, consider this instead.

[Artus]

(Original post by DPLSK)
I would prove the assertion that is a proper subgroup of if and only if is cyclic. I would do this using the subgroup test.

Firstly, show that any cyclic subset of is a subgroup.

Secondly, we use proof by contradiction. Assuming the group is not cyclic then we argue that it must be .

Note: The contradiction here comes from the fact that we must have a 'proper' subgroup of .

I hope that helps.

Darren

PS: Alternatively, consider this instead.

Thanks for your answer. For the link that you gave me, it says "is not very tough to see that adjoining to any cyclic subgroup of order 2 an element
of order 3 or to any cyclic subgroup of order 3 an element of order 2 will yield the whole group S3". But then we are not considering the subgroups generated by adjoining an element from order 2 and another element from another order 2 subgroup. Is it ok if you explain why?

Thanks

[DFranklin]

I'd want some justification about why you can ignore the "2 elements of order 2" case, too. One route: if x, y and e are all distinct and x, y are of order 2, show that xy is distnct from any of e, x, y. So the subgroup generated by x and y is of size at least 4. Then by Lagrange...

[Artus]

(Original post by DFranklin)
I'd want some justification about why you can ignore the "2 elements of order 2" case, too. One route: if x, y and e are all distinct and x, y are of order 2, show that xy is distnct from any of e, x, y. So the subgroup generated by x and y is of size at least 4. Then by Lagrange...

Thanks a lot for answering.

But I still have a question. So if the subgroup is {e, x, y, xy} and has 4 distinct elmenents, then it cannot be a subgroup by Lagrange's Theorem. BUT what if xy=x or xy=y? How can we be sure that xy or yx is not equal to x or y or e...?

[DFranklin]

Show xy isn't e, then show xy isn't x or y.

[DPLSK]

(Original post by Artus)
Thanks for your answer. For the link that you gave me, it says "is not very tough to see that adjoining to any cyclic subgroup of order 2 an element of order 3 or to any cyclic subgroup of order 3 an element of order 2 will yield the whole group S3". But then we are not considering the subgroups generated by adjoining an element from order 2 and another element from another order 2 subgroup. Is it ok if you explain why?

Thanks

I think you can see just by inspection that you'd lose the identity element in every case, something which you don't want to happen.

Darren

[Artus]

(Original post by DPLSK)
I think you can see just by inspection that you'd lose the identity element in every case, something which you don't want to happen.

Darren

But I woudn'tloose the identity if I chose, say, g and gf since g^2=identity

[DPLSK]

(Original post by Artus)
But I wouldn't lose the identity if I chose, say, g and gf since g^2=identity

Consider the following.

where and are two 2-cycles (which do not give the identity).

If then we simply get .

If then we lose the identity.

This is because the identity is mapped to and is mapped to .

doesn't give the identity since .

I hope this helps.

Darren