# Thread: Normal subgroup of S4

1. ## Normal subgroup of S4

Here is the problem:
Let N < S4 consist of all permutations (sigma) such that (sigma)(4) = 4. Is N normal in S4?
I get that
N = {(1234), (2134), (3214), (1324), (2314), (3124)}
(Where
$(2134) = \binom {1 2 3 4}{2 1 3 4}$
Sorry, I don't do well with cyclic notation.)

I know I can do this "brute force" by forming all the left and right cosets. I'd like to find a less time consuming method, but haven't found any yet. Any suggestions?

-Dan

2. Originally Posted by topsquark
Here is the problem:
I get that
N = {(1234), (2134), (3214), (1324), (2314), (3124)}
(Where
$(2134) = \binom {1 2 3 4}{2 1 3 4}$
Sorry, I don't do well with cyclic notation.)

I know I can do this "brute force" by forming all the left and right cosets. I'd like to find a less time consuming method, but haven't found any yet. Any suggestions?

-Dan
Look at the orbit of (123). In general, two elements of S_n are conjugate if and only if their disjoint cyclic notations look' the same.

Sorry - just realised - your N is wrong. It should consist of all permutations where 4 does not appear in it's disjoint cycle notation! You can think of it as all the elements of S_4 written in disjoint cycle notation.

So, N={id, (12), (13), (23), (123), (132)}, which is isomorphic to S_3 (obviously).

3. Originally Posted by Swlabr
Look at the orbit of (123). In general, two elements of S_n are conjugate if and only if their disjoint cyclic notations look' the same.

Sorry - just realised - your N is wrong. It should consist of all permutations where 4 does not appear in it's disjoint cycle notation! You can think of it as all the elements of S_4 written in disjoint cycle notation.

So, N={id, (12), (13), (23), (123), (132)}, which is isomorphic to S_3 (obviously).
Actually I believe my N is correct, it is my notation that is confusing you. Explicitly and in the order I listed them originally I have the group elements:
$\binom{1234}{1234}$

$\binom{1234}{2134}$

$\binom{1234}{3214}$

$\binom{1234}{1324}$

$\binom{1234}{2314}$

$\binom{1234}{3124}$

As far as cyclic notation is concerned I know all of your elements except the last one. (132) is the kind of notation that has never made sense to me.

In any event "In general, two elements of S_n are conjugate if and only if their disjoint cyclic notations look' the same." I can't think of what you mean by this. Are you saying something along the lines of (123) and (1234) are conjugate? If not could you show me an example of two conjugate elements?

-Dan

4. in your N, σ(4) = 4, in other words σ fixes 4. therefore, we can identify σ with an element of S3 (by just ignoring what each σ does to 4, because it doesn't ever do anything to 4).

cycle notation is more compact:

(1 2 3 4)
(2 3 1 4) and

1-->2
2-->3
3-->1
4-->4

both explicitly define some σ by listing the domain and co-domain elements, but these are unwieldy, especially for σ in Sn, where n is large.

one way is to list just the image set {σ(1),σ(2),σ(3),σ(4)} as you have done. but this can easily be confused with cycle notation, which you really

ought to learn. in cycle notation, the permutation listed above is: (1 2 3). this means 2 is the image of 1 (σ(1) = 2), 3 is the image of 2

(σ(2) = 3), and 3 "cycles back" to 1 (σ(3) = 1). any element not listed is considered fixed by σ (so 4 is fixed, because it doesn't appear in the cycle).

the reason cycle notation is really super is this: we can write a conjugate of a cycle as σ(a1 a2 ..... ak)σ^-1 = (σ(a1) σ(a2) ... σ(ak)).

(imagine σ(aj) = rj. then σ^-1(rj) = aj so (a1 a2 ..... ak)σ^-1(rj) takes rj-->a(j+1) so

σ(a1 a2 ..... ak)σ^-1 takes rj-->r(j+1), that is σ(a1 a2 ..... ak)σ^-1 takes σ(aj) to σ(a(j+1)) which is what

(σ(a1) σ(a2) ... σ(ak)) does. you have to treat the case of ak seperately, but that isn't too hard:

σ^-1(σ(ak)) = ak, so σ(a1 a2 ... ak)σ^-1(σ(ak)) = σ((a1 a2 ... ak)(ak)) = σ(a1). if

b is some element of {1,2...,n} that isn't one of the σ(aj), then σ(a1 a2 ... ak)σ^-1(b) = σ(σ^-1(b)) = b.)

to give an example of conjugates in S4, the following two permutations are conjugate:

(1 2)(3 4) and (1 3)(2 4), where (1 2)(3 4) =

1-->2
2-->1
3-->4
4-->3, or what you would write as {2,1,4,3} (1 and 2 change places, and 3 and 4 change places), while (1 3)(2 4) =

1-->3
2-->4
3-->1
4-->2, or what you would write as {3,4,1,2} (1 and 3 change places, 2 and 4 change places).

the advantage of cycle notation is that in it, it is immediately apparent that (1 2)(3 4) and (1 3)(2 4) are conjugates.

your original question can be rephrased like so: is there any conjugate of an element in N that is not in N?

if one accepts that all 3-cycles in S4 are conjugate, this means: is there a 3-cycle in S4 that changes 4?

and of course there is, (1 4 2) is one such 3-cycle. (you would write:

1-->4
2-->1
3-->3
4-->2, or {4,1,3,2}. perhaps you can see how your way of expressing elements in S4 obscures the behavior of the permutations.

it is not immediately apparent that {4,1,3,2} fixes 3, for example, or that it moves 4.

but i have not proved that any two k-cycles for k ≤ n are indeed conjugate. and here, we do not need to do so,

we just need to find SOME element σ in S4, with σ(1 2 3)σ^-1 = (1 4 2). if you followed the argument above,

you can see that any σ in S4 with:

σ(1) = 1
σ(2) = 4
σ(3) = 2 will do. so, for example, we could choose σ = (2 4 3), or:

1-->1
2-->4
3-->2
4-->3

doing the composition long-hand, we have σ{2,3,1,4}σ^-1 as follows:

1-->1-->2-->4
2-->3-->1-->1
3-->4-->4-->3
4-->2-->3-->2, which is (1 4 2), or in your notation {4,1,3,2} which obviously is not in N, so N is not normal.

5. Thanks Swlabr and Deveno. I'll have to take this last post apart (but not tonight), so it might take me a bit to reply.

-Dan

6. Originally Posted by topsquark
In any event "In general, two elements of S_n are conjugate if and only if their disjoint cyclic notations look' the same." I can't think of what you mean by this. Are you saying something along the lines of (123) and (1234) are conjugate? If not could you show me an example of two conjugate elements?

-Dan
Ah - sorry, I should have given an example. Deveno covers this in his post where he says that if a is some permutation, then a(x y...z)a^{-1} = (xa ya...za). That is, you stick the number* x into a and see what you get, y into a and see what you get, ..., z into a and see what you get.

So, by this, an i-cycle (a cycle of length i) is conjugate to a j-cycle if and only if i=j.

From this, you can prove that if an element in disjoint cycle notation is c_1c_2...c_m, and it is conjugate to d_1d_2...d_k, then m=k and c_i is conjugate to d_i for all i (up to permuting the d_i). The converse also holds - if you have two elements c_1c_2...c_m and d_1d_2...d_m with c_i conjugate to d_i for all i, then c_1c_2...c_m is conjugate to d_1d_2...d_m.

Basically, in S_6 (so you'll get more of a flavour for it),

-all elements of the form (ab) are conjugate,
-all elements of the form (ab)(cd) are conjugate,
-all elements of the form (ab)(cd)(ef) are conjugate,
-all elements of the form (ab)(cde) are conjugate,
-all elements of the form (ab)(cdef) are conjugate,
-all elements of the form (abc) are conjugate,
-all elements of the form (abc)(def) are conjugate,
-all elements of the form (abcd) are conjugate,
-all elements of the form (abcde) are conjugate,
-all elements of the form (abcdef) are conjugate,

for all a, b, c, d, e, f in {0, 1, 2, 3, 4, 5, 6}.

*x isn't necessarily a number - it can be anything. But people tend to think of the x as numbers. It is, more correctly, an element of the set your permutations are acting on.

7. Originally Posted by Deveno
the advantage of cycle notation is that in it, it is immediately apparent that (1 2)(3 4) and (1 3)(2 4) are conjugates.
I was able to follow this (with difficulty, but I did it) and I have a question. You say that it is "immediately apparent" that (1 2)(3 4) is conjugate to (1 3)(2 4). This implies that there exists a $\sigma$ such that
$\sigma$((1 2)(3 4)) $\sigma^{-1}$ = ( $\sigma$(1) $\sigma$(2))( $\sigma$(3) $\sigma$(4)) = (1 3)(2 4)

$\sigma$ is fairly easy to find. But $\sigma$ is a member of S4 and since that contains all possible combinations of 1, 2, 3, and 4 it would seem to me that we could find such a $\sigma$ for any element of S4 that can be written in the form (a b)(c d). So are all elements (a b)(c d) (what are they called? Two cycles?) conjugate to each other?

-Dan

Edit: Ah, that's what Swlabr was saying. Okay, between the two of you I think I got it. Thank you!

8. yes, any disjoint pair of 2-cycles are conjugate. an interesting side note, the set of 3 possible disjoint 2-cycle pairs, actually forms (along with the identity) a normal subgroup of S4.

the only "not obvious" part of this, is that they actually DO form a subgroup, but any such product of two different disjoint pairs can be written as:

(a b)(c d)(a c)(b d), for a suitable choice of a,b,c,d from {1,2,3,4}.

this product takes a-->c-->d, b-->d-->c, c-->a-->b, and d-->b-->a, so it is (a d)(b c), the remaining pair.

the normality is immediate from their "disjoint cycle decomposition form".

9. Thank you for all your help. And you know the interesting thing? I started this thread hoping to find a quicker way to do the problem than using the 18 other elements of S4 not contained in N. The thing is that apparently I would have found out that N is not normal from the first calculation...

Ah well. I learned alot anyway, so it's all good.

-Dan