0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\sigma$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.