I am studying a theorem right now in group theory which has 10 lemmas, and there is a part that I didn't understand exactly (which was in the first lemma). On the assumption that the first one is correct (almost) the rest of the proof is clear. The theorem goes like this:
$Theorem:$ Suppose that G is a permutation group on letters $c_{0},...,c_{n-1}$ such that:
$1)$ G is doubly Transitive.
$2)$ Only the identity fixes two letters.
$3)$ At most one element taking $c_{i}$ into $c_{j}$ displaces all letters, or
or ($3')$ n is finite.)
Then the identity and the elements of $G$ displacing all letters form a transitive normal group.
The proof mostly concentrates on the set of permutation of order 2. At first it claims that there is one and only one element of order 2 in $G$ that interchanges a specified pair of letters. It's easy to prove that if there exists one then that's the only one, but what promises that there exists such element of order 2? Since G is doubly transitive I know for sure that there exists a permutation with the cycle (i,j), but it's not necessarily or order 2. The second condition says that there is at most 1 letter that is fixed. And if I understand the third property correctly if a permutation send $c_{i}$ to $c_{j}$ (where the two letters are not necessarily different), if the permutation displaces the rest, then that's the only permutation with that property, and as I see it this doesn't really help as well. It seems like there is something trivial that I am missing. Would be happy if someone clarify. Thank you.