2
$\begingroup$

I'm trying to understand actions in regards to group theory . specifically in my notes I found the following example :

Say G=$A_4$, for $x \in \{1,2,3,4\}$, and $\tau \in A_4$

We let $x^{\tau}$ be the usual permutation action, let $\tau=(1,4,2,3)$

Then $1^{\tau}=4,2^{\tau}=3,3^{\tau}=2, 4^{\tau}=2$.

My thought's :

I thought this just meant that we perform the permutation on the set

i.e. $x=\{1,2,3,4\}=\{x_1,x_2,x_3,x_4\}$

then $x^{\tau}$ just permutes these elements according to $\tau$.sending $x_1x_2x_3x_4 \rightarrow x_3x_4x_2x_1$

giving $1^{\tau}=4,2^{\tau}=3,3^{\tau}=1,4^{\tau}=2$

Which almost agrees with the example except for at 3 ? What am I misunderstanding ?

$\endgroup$
  • 2
    $\begingroup$ $3^{\tau}=4^{\tau}$ says that $\tau$ is not injective, a contradiction. $\endgroup$ – Dietrich Burde Nov 4 '18 at 19:56
  • $\begingroup$ I think $3^\tau=2$ was just a typo, and your $3^\tau=1$ is correct. $\endgroup$ – Andreas Blass Nov 5 '18 at 1:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.