# $G \leq S_n$ transitive and n prime $\Rightarrow$ $n$ divides $ord(G)$

I am doing a task and therefore I need to show the following:

$G \leq S_n$ transitive and n prime $\Rightarrow$ $n$ divides $ord(G)$

I think it's quite obvious, but I don't know, how to proof it formally right.

Transitivity means that for all $x,y \in G$ exists $g \in G$ such that $x^g=y$, where $x^g$ denotes the conjugation $g^{-1}xg$.

• Your definition of transitive is not correct. Your definition seems to be for a group to act transitively on itself by conjugation. But this is not possible, because given $x$ there are $|G| -1$ choices for $y \neq x$. But $x^{e} = x^{x} = x$, so there are at most $|G|-2$ possibilities for $x^{g}$. – Morgan Rodgers Nov 21 '17 at 22:48
• When you talk about a group being transitive, your first question should be "what set is it acting transitively on, and how is it acting?" In this case, "on itself, by conjugation" is not an answer that works. – Morgan Rodgers Nov 21 '17 at 22:54
• Ahh, so it's acting transitive on $S_n$. – Myrkuls JayKay Nov 22 '17 at 9:00
• It’s also not possible, if it is a subgroup of $S_n$, to be acting transitively on the elements of $S_n$ by conjugation. It’s probably meant to be acting on $\{1,\ldots n\}$ in the natural way (as permutations) – Morgan Rodgers Nov 22 '17 at 16:35
• Yes, saying that "$G\le Sym(X)$ is transitive" means that $G$ acts transitively on the set $X$. Here $X=\{1,2,\ldots,n\}$ as Morgan Rodgers suggested. – Jyrki Lahtonen Jan 21 '18 at 18:22

If $G$ is transitive on $n$ elements, then $G$ has a subgroup of index $n$ by the orbit stabilizer theorem, so $n$ divides the order of $G$.
• So "$n$ prime" is not needed? – Hagen von Eitzen Nov 21 '17 at 22:43
• If $G$ is transitive on $n$ elements, say $\lbrace 1,2 \cdots n \rbrace$, then $|Orb(1)|=n$ and hence $[G:Stab(1)]=|Orb(1)|=n$ – mich95 Nov 21 '17 at 22:48
• My aim in the end is to show, that it exists a subgroupb $H$ of $G$ such that $ord(H)=n$. Therefore my first idea was to use Cauchys Theorem, so I need that $n$ divides $ord(G)$. It is possible that I need the prime property of $n$ only there. But how do I know, that $G$ is transitive on $n$ elements? – Myrkuls JayKay Nov 21 '17 at 22:50
It is impossible for a group $G$ to act transitively on itself by conjugation. To see this, notice that if you take the identity element $e$, you have $e^{g} = e$ for all $g \in G$. So most likely, if you found this problem in a book, what is meant is that since $G$ is a subgroup of $S_{n}$, $G$ is a collection of permutations of $\{1,2,\ldots,n\}$ (so the action is simply by permuting these numbers).
To say $G$ is transitive means that for any $i,j \in \{1,\ldots,n\}$, there is a $g$ in $G$ with $i^{g} = j$, where $i^{g}$ means you apply the permutation $G$ to $i$ in the natural way. Then you can apply the orbit-stabilizer theorem to show that $n$ divides $|G|$. You don't need the fact that $n$ is prime.