Thread: Symmetric Group Subgroups

Find all subgroups of and determine which are normal.

I'm not sure if this is correct, but so far I have
{(1,2,3), (2, 3, 1), (3,1,2), (2, 1,3), (1,3,2), (3,2,1)}

what do you mean by (1,2,3)? is this the image of the set (1,2,3) under the map of some element of S3:

1-->1
2-->2
3-->3

or is this supposed to mean the 3-cycle (1 2 3), which is the map:

1-->2
2-->3
3-->1 ?

there are two subgroups of S3 that you should be able to list, without even thinking about it at all.

I meant the image. Two subgroups I can think of quickly are S_3 and I

ok, consider subgroups of the form <x> where x is an element of S3. how many of these are there?

what possible orders can subgroups have, anyway?

I'm a little confused about the notation. I was letting x = (2,3,1) and I found that x^3 =1, so that has order 2. I think then that <x>={1, x, x^2}
How do I know how many there are short of listing all possible matrices and multiplying them?
And I'm not sure what orders are possible in general.

have you covered lagrange's theorem yet?

by the way, if x^3 = 1, and x is not the identity, then the order of x is 3, not 2.

No, I haven't heard of Lagrange's thm.

do you know what the subgroup generated by an element is?

i find it rather incredible that you have gotten all the way to normal subgroups and still not covered lagrange's theorem.

here it is:

if H is a subgroup of a finite group G, then the order of H is a divisor of the order of G.

I know how to check if something is a subgroup, but I don't know how to find a subgroup. I thought that what I had before (<x>={1, x, x^2}) was the subgroup generated by x. Is that wrong?

no, that is correct. the reason i asked about lagrange's theorem, is that it tells you S3 can only have groups of order 1,2,3 or 6.

if a subgroup is of order 6, it has to be S3 itself <---always normal
if a subgroup is of order 1, it must be the identity element {e} <---always normal

that just leaves 2 and 3. how many subgroups of each order can you find....and which of these are normal?

<x> has order 3
If y= (2,1,3) then <y> has order 2
I think xy is also a subgroup with order 2.
I can't think of any others. I'm not even sure these are right.

what you need, my friend, is a better notation, so we can talk about these elements of S3 more easily.

there does exist one, it is called cycle notation. so instead of (2,1,3), which means 1 goes to 2, 2 goes to 1, 3 goes to 3, we can write:

(1 2) which means: 1 goes to 2, 2 goes to 1, 3 stays the same (not part of "what changes").

in this notation your (2,3,1) becomes (1 2 3) ("1 goes to 2, 2 goes to 3, 3 goes to 1").

you are correct <x>= <(1 2 3)> has order 3, and <y> = <(1 2)> has order 2.

you are also correct that <xy> is also of order 2. there is one more subgroup of order 2 you have missed.

now, can you tell me why <x^-1> = <(1 2 3)^-1> = <(1 3 2)> does not give us a new subgroup?

after all, x^-1 should be of order 3 as well.

We never use cycle notation in my class, and it seems like everything I find when studying uses it. I don't know why she doesn't teach it. I'll give it a try:

<yx^2>={(2,3), I} is the group I forgot.

I think in this group, x^-1 = x^2 because x^3 =I, so x^-1 is already counted.

To check if they're normal, do I need to go through every element of S_3 and check?
I mean check <x>: x*x*x^-1, y*x*y^-1, x^2*x*x^-1, xy*x*(xy)^-1, yx^2*x*(yx^2)^-1
and do that for all elements of <x> and <y> and <xy> and <yx^2>? Seems like a lot of work...

proving "not-normal" is sometimes easier than proving normal.

if you find some g in G for which gHg^-1 ≠ H, you can stop looking, H is not normal.

in S3, there is one normal subgroup that is proper and non-trivial.

So <x> is the only normal proper subgroup.