Find all subgroups of $\displaystyle S_3$ 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)}

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- Oct 7th 2011, 06:30 PMveronicak5678Symmetric Group Subgroups
Find all subgroups of $\displaystyle S_3$ 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)}

- Oct 7th 2011, 07:08 PMDevenoRe: Symmetric Group Subgroups
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. - Oct 7th 2011, 07:25 PMveronicak5678Re: Symmetric Group Subgroups
I meant the image. Two subgroups I can think of quickly are S_3 and I

- Oct 7th 2011, 07:48 PMDevenoRe: Symmetric Group Subgroups
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? - Oct 7th 2011, 07:56 PMveronicak5678Re: Symmetric Group Subgroups
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. - Oct 7th 2011, 07:57 PMDevenoRe: Symmetric Group Subgroups
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. - Oct 7th 2011, 08:01 PMveronicak5678Re: Symmetric Group Subgroups
No, I haven't heard of Lagrange's thm.

- Oct 7th 2011, 08:24 PMDevenoRe: Symmetric Group Subgroups
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. - Oct 7th 2011, 08:49 PMveronicak5678Re: Symmetric Group Subgroups
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?

- Oct 7th 2011, 10:53 PMDevenoRe: Symmetric Group Subgroups
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? - Oct 8th 2011, 09:44 AMveronicak5678Re: Symmetric Group Subgroups
<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. - Oct 8th 2011, 09:53 AMDevenoRe: Symmetric Group Subgroups
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. - Oct 8th 2011, 10:16 AMveronicak5678Re: Symmetric Group Subgroups
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... - Oct 8th 2011, 10:54 AMDevenoRe: Symmetric Group Subgroups
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. - Oct 8th 2011, 11:16 AMveronicak5678Re: Symmetric Group Subgroups
So <x> is the only normal proper subgroup.