# Normal groups

1. Nov 8, 2008

### Marin

Hi!

I want to determine which of the subgropus of the symmetric group S(3) are normal. The condition is:

for every g in G: g H g^-1 = H where H is a subgroup of G.

I have determined all the subgroups of S(3) and I came up with 13.

What I did after that is I considered 2 cases:

1: g is elem. of H, then g^-1 is also in H and the condition is satisfied.
2: g is not el. of H, but still el. of G, the same is true for g^-1.

then I got stuck :(

theoretically all the subgroups could be examined one by one, but it seems to me somehow too long and "not mathematical"

Does anyone know a better way?

° and another question: If H is a normal subgroup of G, what should one understand under "G modulo H"

2. Nov 8, 2008

### sutupidmath

well, as you know:

$$S_3=\{(1),(12),(13),(23),(123),(132)\}$$ Now since $$o(S_3)=6$$ it follows that its subgroups can be only of order 1, 2, 3 or 6. OF course the only subgroups of ord. 1 and 6 are the trivial one and S_3 itself respectively.

So we are left with proper subgroups of ord 2 and 3. which are

$$H_1=\{(1),(12)\}, H_2=\{(1),(13)\},H_3=\{(1),(23)\},H_4=\{(1),(123),(132)\}$$

So where did you get 13 subgroups?

Then what i would do here, since the order of S_3 is not large, just look at the table for the operaton of S_3, and determine for which subgroups H_i, i=1,2,3,4=> gH=Hg. for every g in G.

H_1, is not normal since, $$(123)^{-1}(12)(123)=(132)(12)(123)=(1)(23)=(23)$$ thus (23) is not in H_1.

3. Nov 9, 2008

### Marin

hmmm in my book the 6 elements are denoted:

(123), (132), (213), (231), (312), (321) - how did you transform them in (1), (12), (13), (23), (123), (132)

4. Nov 9, 2008

### sutupidmath

well what does (123) mean??

It simply means the following, let f be the mapping then: f: 1-->2; 2-->3, 3-->1

so following this logic, it yields that

(123)=(231) =(312)and also (132)=(321)=(213) and these are actually only your even permutations in S_3, including (1) of course since (1)=(12)(21)

Well, either you are misinterpreting what it is written in the book, or it is a typo, because the 6 elements of the symmetric group S_3 are the ones i listed.

it is simply all the mappings that go from f:X_3--->X_3 where X_3={1,2,3} such that f is bijective.

5. Nov 9, 2008

### Marin

thanks, sutupidmath, I think I got it

could you also help me eoth this: "If H is a normal subgroup of G, what should one understand under "G modulo H" ". I just cannot figure out what it means G modulo H in this context.

6. Nov 9, 2008

### HallsofIvy

If your textbook is defines normal subgroups it certainly should have a definition of G/H!

A subgroup, H, of a group G is normal if and only if the right cosets, Hx, are the same as the left cosets, xH. That's important because it allows us to put an operation on the cosets: If xH and yH are two different left cosets of H, then "(xH)(yH)" is the left coset (xy)H. Of course, z is another member of xH, we can also represent that "zH". Would (zH)(yH)= (zy)H give the same coset? It can be shown that it will if and only if H is a normal subgroup of G- that the left and right cosets are the same.

In that case, we define G/H to be the collection of all left cosets of H with that operation as group operation.

7. Nov 9, 2008

### Marin

but if G/H is a subgroup of G, for that xH=Hx for every x of [x], could we say then that G/H Abelian is?

8. Nov 9, 2008

### HallsofIvy

No, that does not follow.

9. Nov 9, 2008

### Marin

well I think I have to take a break now :)

Thank you very much for the help!