Question

Need help understanding commutator subgroup
Please

Answer 1

@perl

Answer 2

calculate the commutator subgroup of S_3

Answer 3

\(S_3= \{(12),(13),(23),(123),(132)\}\)
and commutator in \(S_3\) must be an even permutation. So that, let H is commutator subgroup of \(S_3\) , then H=\(\{(12),(13), (23)\}\)
but I have to show that H is the required subgroup

Answer 4

commutator subgroup means the elements are commutative?

Answer 5

you are missing one element in S3, (1) the identity

Answer 6

nope, the definition of commutator subgroup is \(\{aba^{-1}b^{-1}|a, b\in G\}\)

Answer 7

a, b are commutators

Answer 8

you should have six elements in S3

Answer 9

Yes, I listed above

Answer 10

check again, you left out (1)

Answer 11

oh yes, identity. (123)

Answer 12

the identity is just (1)

Answer 13

the same

Answer 14

not a big deal , though

Answer 15

(123) is not the same as (1) , right?

Answer 16

literally, no. However, when talking about identity, I indicate to \(\left(\begin{matrix}1&2&3\\1&2&3\end{matrix}\right)\)
so that, mathematically, it is 1. hehehe

Answer 17

right

Answer 18

this is kind of advanced for me, do you have notes on this, i can read

Answer 19

http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Commutator_subgroup.html

Answer 20

on the commutator stuff

Answer 21

so you found
[(1,2),(2,3)] ?

Answer 22

it might help to identify the inverses of these elements. let me see if i can find a table for S3

Answer 23

here http://groupprops.subwiki.org/wiki/Symmetric_group:S3

Answer 24

so
(1,2)^-1 = (1,2)
(2,3)^-1 = (2,3)
(1,3)^-1 = (1,3)
(1,2,3)^-1 = (1,3,2)
(1,3,2)^-1 = (1,2,3)

Answer 25

how?? how (1,2) ^-1 = 1,2?

Answer 26

because (1,2)(1,2) = ( )

Answer 27

if you think of S3 as the motions of an equilateral triangle, ( 1 , 2 ) is a reflection about an axis. so if you do the reflection again, you are back to where you start

Answer 28

oh, yes. My goodness.