# Transitive group action and meaning of "The usual action of $S_n$"

I ran into an example of a transitive group action. It stated: For $n$ $\ge$ $1$, the usual action of $S_n$ on $\{1, 2, . . . , n\}$ is transitive since there is a permutation sending $1$ to any number. $$\\$$ My questions are basic - firstly, what does "the usual action of $S_n$ on $\{1, 2, . . . , n\}$" mean? Secondly, I know the definition of transitive group action but I'm pretty sure I do not totally understand it. I would like to understand why the above example is an example of a transitive group action (rigorous explanation of what ties the definition of transitive action with this action). $$\\$$ Thank you.

• $S_n$ is by definition the group of permutations of $\{1,2,\dots,n\}$, so the usual action means each $\sigma\in S_n$ acts by $\sigma\cdot i=\sigma(i)$. Dec 12 '16 at 23:03
• I know what $S_n$ is, Guess I forgot to mention it. And - OK got it (this answers my first question). Thanks. Dec 12 '16 at 23:07

Isomorphic groups can act differently on different sets. For instance the dihedral group of order 8 $D_8$ can act on eight points (a flat square box) giving as generators the elements $(1,2,3,4)(5,6,7,8)$ (a 90* rotation) and $(1,5)(2,6)(3,7)(4,8)$ (flippping the box inside out). It also has an action on a square with rotation $(1,2,3,4)$ and a front-back swap $(1,2)(3,4)$. The number of points do not even have to differ. The Klein Vierergruppe ($C_2 \times C_2$) has an action on four points with even permutations : $\{(), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)\}$ and another one containing odd permutations : $\{ (), (1,2), (3,4), (1,2)(3,4)\}$. While some groups can have transitive actions they also can have non transitive ones. The dihedral action on the square is transitive but its action on pairs of points is not, adjacent pairs are never mapped to opposite ones.
• First off, thanks. Let's take the box inside out example. According to the definition of the transitive action, for every $x,y$ there exists a group element $g$ such that $g.x=y$. But if we take $x=1$, $y=6$ we don't have any $g \in G$ such that $g.x=y$. (What am I missing?) Dec 13 '16 at 14:06
• Let $r = (1,2,3,4)(5,6,7,8)$ and $s=(1,5)(2,6)(3,7)(4,8)$ then $rs=(1,6,3,8)(2,7,4,5)$ clearly maps $1$ to $6$, where is the problem? Dec 13 '16 at 14:43
• OK just one more thing - Does this, by definition of orbit, mean $rn$ $\in$ $G.1$ (the orbit of $1$) and $|G.1|$ $=$ $8$? I'm literally crossing my fingers. Dec 13 '16 at 18:57