# A $t$-transitive permutation group of degree $n$.

Suppose we have a $t$-transitive permutation group $G$ of degree $n$, where $t$ is greater or equal to $2$.

Then why does this not imply that $G = S_n$?

As since $G$ acts $2$-transitively then $G$ must contain all transpositions, and therefore in fact be $S_n$.

Where am I going wrong?

The line "since $G$ acts $2$-transitively it must contain all transpositions" is wrong. $G$ being $2$-transitive means that for every $a,b,a',b'\in X$ (the set acted on by $G$) we have some $g\in G$ s.t. $ga=a', gb=b'$. In order for $g$ to be a transposition, it would additionally have to fix every other element of $X$, but this need not be the case.

"As since G acts 2 transitively then G must contain all transpositions" -this is not true.

"Then why does this not imply that $G = S_n$?" - E.g., $A_n (n\ge 3)$ is $(n-2)$-transitive.

Consider also this.

A finite group which is $2$-transitive on a set $X$ with $n$ elements must have order divisible by $n (n-1)$.

Now for a prime $p$ the group of all maps $x \mapsto a x + b$, for $a \in \mathbf{F}_{p}^{*}$ and $b \in \mathbf{F}_{p}$ acts $2$-transitively on $\mathbf{F}_{p}$, and has order exactly $p (p-1)$, that is, as small as possible.

Look at the alternating group $A_n$. It is $(n-2)$-transitive and does not contain any transposition.

If $G$ is 2-transitive, then for all ordered pairs $(i,j),(k,\ell)$, there is an element $g \in G$ such that $g$ takes $i$ to $k$ and $j$ to $\ell$. Thus, $g$ would have the cycle decomposition $g=(i,k,\ldots,j,\ell,\ldots) \cdots$ or $g=(i,k,\ldots)(j,\ell,\ldots) \cdots$. Observe that this $g$ is not a transposition, and so $G$ need not contain any transpositions at all. On the other hand, the following assertion is true: If $G$ is 2-transitive and contains at least one transposition, then it contains all transpositions and hence equals $S_n$. Proof: if $G$ is 2-transitive and contains $(i,j)$, then for any other transposition $(k,\ell)$, there is some $g \in G$ such that conjugation by $g$ gives $g^{-1}(i,j)g = (k,\ell)$, whence $G$ contains $(k,\ell)$ also; since $G$ contains all transpositions, it equals $S_n$.