First of all, this is again one of my questions ! Sorry, at the moment I've got plenty of questions ! There is something I just struggle with since hours. In an exam I got, I had to show that if we have $p$ a prime number and $A$ a set of $p$ elements, then :
$S_p \times A^p \rightarrow A^p$ defined by $(\sigma, (a_1, \dots, a_p)) \rightarrow (a_{\sigma^{-1}(1)}, \dots, a_{\sigma^{-1}(p)}) = \sigma .(a_1, \dots, a_p)$
is a group action. But I don't find out why ! Thus, if I take for example $A = \{a_1, a_2, a_3\}$ a set of three elements and I consider $s=(1 \; 2)$, $t=(1 \; 2 \; 3)$, we have : $t.(a_1,a_2,a_3) = (a_{t^{-1}(1)}, a_{t^{-1}(2)}, a_{t^{-1}(3)}) = (a_3, a_1, a_2)$, and $s.(t.(a_1,a_2,a_3)) = (a_3, a_2, a_1)$. Now if the application is an action of group, we must have : $(st).(a_1,a_2,a_3) = s.(t.(a_1,a_2,a_3))$ ; but we have $st = (2 \; 3)$, so $st.(a_1, a_2, a_3) = (a_1, a_3, a_2) \neq (a_3, a_2, a_1)$. So, I find that finally, this is not a group action !
Is there somewhere I made a mistake ? In the same way :
$s.(t.(a_1, \dots, a_n)) = s.(a_{t^{-1}(1)}, \dots, a_{t^{-1}(p)}) = (a_{s^{-1}(t^{-1}(1))}, \dots, a_{s^{-1}(t^{-1}(p))}) = (a_{s^{-1}\circ t^{-1}(1)},\dots, a_{s^{-1}\circ t^{-1}(p)})$,
and as : $s^{-1} \circ t^{-1} = (ts)^{-1}$, we have : $s.(t.(a_1, \dots, a_n)) = (ts).(a_1, \dots, a_p)$.
But I must have made a mistake, cause I have to show that this is an action... But I don't find where !
Someone could help me ?
Observe that when you make act $\;(12)\;$ on something, it is in the order they are given, thus:
$$s(t(a_1,a_2,a_3)=s(a_3,a_1,a_2)=(a_1,a_3,a_2)$$
and now $\;st=(23)\;$ , and thus
$$st(a_1,a_2,a_3)=(a_1,a_3,a_2)=s(t(a_1,a_2,a_3)) ...!$$
If it is not with the above agreement, what you're given is, certainly, not a group action...