Let $G$ be a transitive subgroup of $S_p$ where $p$ is an odd prime number. Now consider the following assumptions -

$(i)$ $G$ is solvable.

$(ii)$ If $\sigma \in G$ and there exist $h\ne j$ such that $\sigma(h)=h$ and $\sigma(j)=j$, then $\sigma=\mathrm{id}$.

It is true that $(i)$ implies $(ii)$. Is the converse true?

Edit. Thank you for the answear. I already knew that $(i)$ implies $(ii)$.

Here's the original problem I was solving - given $f\in \mathbb{Q}[x]$ - irreducible polynomial whose degree is a prime number, prove that $f(x)=0$ is solvable in radicals if and only if the splitting field of $f$ could be generated by any two of its roots.

My ideas. If the equation is solvable, then knowing the result $(i)\Rightarrow(ii)$ one observes that $\text{Gal}(K,\mathbb{Q}[\alpha,\beta])=\{\mathrm{id}\}$, where $K$ is the splitting field of $f$ and $\alpha,\beta$ are some of the roots.

For the opposite direction we see that if $K=\mathbb{Q}[\alpha,\beta]$ for any two roots of $f$, then $(ii)$ holds. So my idea was to use the converse, if it is true. It is not, as you say, but however we could notice that $[\mathbb{Q[\alpha,\beta]}:\mathbb{Q}]\le p(p-1)$ ($p$ is the degree of the polynomial). Thus $\text{Gal}(f)$ has only one Sylow p-subgroup which is therefore normal. Hence $\text{Gal}(f)$ is contained in the normalizer of a Sylow p-subgroup in $S_p$, which according to a problem in my book is a special group of order $p(p-1)$ which is solvable. Hence $\text{Gal}(f)$ is solvable.


Any transitive solvable subgroup of $S_p$ is conjugate to a subgroup of the group of affine linear transformations. That means if $\sigma \in G$ then $\sigma = ax + b$, $a \in (\mathbb{Z}/p\mathbb{Z})^*$, $b\in \mathbb{Z}/p\mathbb{Z}$. Now suppose we have a $\sigma$ as in the hypothesis. Then $h = ah + b$, $j = aj + b$. Subtracting gives $h-j = a(h-j)$, forcing $a =1$ since $h \neq j$. So $\sigma =$ id.

The converse is not true. Here's the idea behind finding a counterexample. Take $G=A_p$, $p$ a large prime. We know $A_p$ is not solvable. Let $\sigma \in A_p$ be a non-trivial permutation of small length and then it should be easy to find elements fixed by $\sigma$.


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