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)}
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'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.
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.
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?
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...