2

I've been having problems with finding conjugacy classes. I don't really understand how to do it properly.

Say we look at a S3 group: $S_3=]e, (12), (13), (23), (123), (132)]$

If we look at just $(12)$ first:

$e(12)e^{-1} = e(12)e=(12)$ I can understand this.

$(12)(12)(12)^{-1} = (12)(12)(12) = (12)$ Now here I'm a little confused. Wouldn't $(12)^{-1} = (21)$?

Then $(13)(12)(13)^{-1} = (13)(12)(13) = (32)$ how did that become $(32)$?

I thought you would start from the right so do $(12)(13)$ first then that would be $(123)$ then $(13)(123)$: 1 stays the same. 3 goes to 2? then 3 stays so (123)? I'm really confused

Kennan
  • 351
  • $(12) = (21)$. They're both the permutation that switches $1$ and $2$. – Jair Taylor Jun 13 '15 at 06:04
  • You need to be able to recognize when two written cycles represent the same permutation. For instance, (12345), (23451), (34512), (45123), (51234) are all the same permutation! – anon Jun 13 '15 at 06:28

3 Answers3

1

(12) and (21) are the same permutation.

As for (13)(12)(13), well, simply compute what it does to 1,2,3:

  • (13)(12)(13)1=(13)(12)3=(13)3=1
  • (13)(12)(13)2=(13)(12)2=(13)1=3
  • (13)(12)(13)3=(13)(12)1=(13)2=2

So (13)(12)(13) leaves 1 fixed, and swaps 2 and 3.

Also, (12)(13) is not (123). Check:

  • (12)(13)1=(12)3=3
  • (12)(13)2=(12)2=1
  • (12)(13)3=(12)1=2

Thus, (12)(13) sends 1 to 3, and 3 to 2, and 2 to 1. So (12)(13)=(132), not (123).

anon
  • 155,259
1

$(12)$ and $(21)$ are the same permutation (which maps $1$ to $2$, $2$ to $1$, and leaves everything else unchanged).

Likewise, $(13)(12)(13) = (23)$ because it maps $2$ to $3$, $3$ to $2$ and leaves everything else unchanged (including $1$). Also, $(12)(13) = (132)$, not $(123)$.

fkraiem
  • 3,169
  • Then what about $(123)^{-1}$? I would say that would be $(321)$ but the answer is $(132)$ So the inverse is not the inverse of the permutation?? – Kennan Jun 13 '15 at 06:12
  • 1
    $(321)$ and $(132)$ are the same permutation. – fkraiem Jun 13 '15 at 06:27
1

Your confusion is coming from the composition of permutations. Here is an example suppose we want to know $(1 \, 2 \, 3)( 1\, 3)$. Then we start from the rightmost cycle, i.e. $(1 \, 3)$. Here $1$ goes to $3$, then we consider the next cycle to the left, here $3$ goes to $1$, so we will conclude that the $1$ (from the rightmost cycle) goes to $1$ (because $1 \to 3 \to 1$). Now we consider $3$ from the rightmost cycle, this $3$ goes to $1$, then in the left cycle $1$ goes to $2$, thus $3$ goes to $2$. So we have $$(1 \, 2 \, 3)( 1\, 3)=(1)(3 \, 2)=(3 \, 2).$$ Once you get this then conjugacy will become easier to handle.

Anurag A
  • 42,261
  • Do we always start from the right or it doesn't matter? – Kennan Jun 13 '15 at 06:14
  • 1
    @Kennan think of it as function composition $f(g(x))$. We will start with $g$ and then $f$, likewise we have to start from the rightmost and then work towards left. – Anurag A Jun 13 '15 at 06:16
  • so let's see if I've got this right... if it's for $(13)(132)$, starting from the right cycle. $1->3->2$ then we look at left cycle, there's no 2 so 2 remains? or do we go $1->3->2->1$ then from left cycle $1->3$ but then I'll still need a 2 somewhere... – Kennan Jun 13 '15 at 06:34
  • @Kennan (13)(132)1=(13)3=1 and (13)(132)2=(13)1=3 and (13)(132)3=(13)2=2. So (13)(132) sends 1 to 1, 2 to 3, and 3 to 2. Thus (13)(132)=(23). – anon Jun 13 '15 at 06:36
  • okay one more attempt: $(23)(12)(23)^{-1} = (23)(12)(23)$ then $(23)(12)(23)1=(23)(12)1=(23)2=3$, $(23)(12)(23)2=(23)(12)3=(23)3=2$ and $(23)(12)(23)3=(23)(12)2=(23)1=1$ so $(23)(12)(23)=(13)$. Is that right? – Kennan Jun 13 '15 at 06:47
  • @Kennan Yes that's right. – anon Jun 13 '15 at 07:40