$T \unlhd G \unlhd K$, plus extra conditions, implies $T\unlhd K$?

Question

Consider the symmetric group $S_p $, $p $ prime, and the subgroup $T=\langle (12\dots p) \rangle$. Then consider two subgroups of $S_p $, say $G $ and $K $, such that $T \unlhd G \unlhd K$ and both $G/T $ and $K/G $ are cyclic of prime order. How can I prove that $T $ is normal in $K $?

Since $T $ is not normal in $S_p $, there is (at least) a permutation, say $\sigma $, such that $\sigma T \sigma ^{-1} \ne T$. Clearly $G/T $ can't be generated by $\sigma$ (in that case $T $ wouldn't be normal in $G $, so the notation $G/T $ doesn't even make sense), so maybe I should prove that neither $K/G $ can. Is it right? Even if it is, I have no idea on how to proceed. Thanks for any suggestion

I had this idea: we studied that $T $ is the only $p$-group of $A =\{\sigma: \sigma (x) = ax+b\}$, whose elements are permutations of $\mathbb Z /p \mathbb Z$, with $a\ne 0$. Now we can say that $A $ is a set of all the permutations $\sigma $ such that, if we know $\sigma (x) $ and $\sigma (y) $ for two distinct $x,y\in \mathbb Z /p \mathbb Z$, we can univocally determine $\sigma \in A$. If I prove that $A $ is the only transitive subgroup of the symmetric group with this property, $T $ would be the only $p $-group (since clearly a $p $-cycle has the property above). Is that a good method? I can't show that $A $ is unique anyway

Answer 1

This is basically the same answer as @Yuval, but I think about it a bit differently.

Since $T$ has order $p$ and lives in $S_p$, we know that $T$ is a Sylow $p$-subgroup of $K$. Let $n$ be the number of conjugate subgroups of $T$ in $K$, i.e., $n$ is the number of Sylow $p$-subgroups in $K$. Now consider the action of $K$ on these subgroups, which gives a homomorphism $\varphi\colon K\to S_n$. If we can show that $\ker\varphi=K$, i.e., that this action is trivial, then it follows that $T$ is normal in $K$.

Since $T$ is normal in $G$, $\ker\varphi$ contains $G$. Since $$q=[K:G]=[K:\ker\varphi][\ker\varphi:G]$$ for some prime $q$, we see that $[K:\ker\varphi]$ is either $1$ or $q$. If $[K:\ker\varphi]=1$, then $K=\ker\varphi$ and we are done. If $[K:\ker\varphi]=q$, then by Sylow theory, $q\equiv 1\pmod p$. However, $q$ is a prime less than or equal to $p$, so this is impossible.

Answer 2

Let $S$ be the set of all subgroups of order $p$ in $K$ that are conjugate to $T$. Now $K$ acts on $S$ by conjugation: for any $x\in K$ , it maps $H\in S$ to $xHx^{-1}\in S$.

By assumption, the action of any $x\in G$ on $T$ is trivial. We now argue by contradiction, if $T$ is not normal in $K$, then since $[K: G]$ is prime, $G$ is the isotropy group and $S$ contains $[K:G]$ many elements.

Now we apply Sylow theorem No. 2, which says all subgroups of order $p$ are conjugate to $T$, therefore there are $|S|=[K:G]$ many such $p$-groups. Then by Sylow theorem No.3, $|S|\equiv 1$ mod $p$. But by assumption, $|S|=[K:G]$ is a prime strictly smaller than $p$ (here note the order of $S_p$ contains only one power of $p$), so this congruence relation cannot hold, that is a contradiction.