Theorem about non-regular group action, where an element fixes no points or exactly $p$ points

I have a question on the following proof. All groups are assumed to be finite. But first I will mention a lemmata:

Lemma: Let $G$ act faithfully and non-regular as a group such that there exists some element fixing no point, and every element fixing some point fixes exactly $n$ points. Then $|\Omega| \equiv n \pmod{|G_{\alpha}|}$.

Now I know as a somehow standard result, that if $G$ acts transitively and $N \unlhd G$ is a normal subgroup, then the orbits of $N$ form a system of blocks for $G$. Further a group $A \le G$ is called a t.i. subgroup if it intersects trivially with its conjugates, i.e. $$A \cap A^g = 1 \quad \mbox{for} \, g \notin N_G(A).$$

Fact: If $G$ acts transitively and non-regulary on $\Omega$ as group such that some element fixes no point, and each element fixing some point fixes exactly $n$ points, and if $\alpha \in \Omega$, then $|N_G(G_{\alpha}) : G_{\alpha}| = n$ if and only if $G_{\alpha}$ is a t.i. subgroup.

The Theorem is the following:

Theorem: Let $p$ be a prime and $G$ a faithful non-regular transitive group on $\Omega$ such that there exists some element fixing no point, and every element fixing some point fixes exactly $p$ points. Assume that $G_{\alpha}$ is t.i. for some $\alpha \in \Omega$ and let $\overline \alpha := \mbox{fix}(G_{\alpha})$ be the set of fixed points of $G_{\alpha}$. Let $N$ be a normal subgroup of $G$ of order coprime to $p$ and let $\Sigma$ be the set of orbits of $N$. Then

(a) no orbit of $N$ contains more than one element of $\overline a$,

(b) $G$ acts on $\Sigma$ as a non-regular group such that there exists some element fixing no point, and each element fixing some point fixes exactly $p$ points,

(c) if $|\Sigma| = p$, $G$ has a normal subgroup of index $p$ that acts as a Frobenius group on each of its $p$ orbits,

(d) if $|\Sigma| > p$, $N$ acts semi-regularly on $\Omega$ and $G / N$ acts faithfully on $\Sigma$.

Proof: As $p \nmid |N|$ and $p \mid |\Omega|$, the number of orbits of $N$ is divisible by $p$. Let $\Delta$ be the orbit of $N$ containing $\alpha$. Since $\overline \alpha$ is a minimal $G$ block, either $\Delta$ contains the whole of $\overline \alpha$ or it contains no further elements of $\overline \alpha$. The first cannot occur, because it would imply that $p \mid |\Delta|$. This proves (a). Further from the above Lemma it follows that for any orbits $\Delta'$ of $N$, $|\Delta'|$ is coprime to $|G_{\alpha}|$; so if an element of $G_{\alpha}$ fixed $\Delta'$, if fixes an element of $\Delta'$. Now $G$ acts transitively on $\Sigma$ and by Lagrange's theorem the stabilizer of $\Delta$ is $G_{\alpha}N$. Thus if an element fixed an orbit of $N$ it either fixes all orbits or it fixes exactly $p$ of them. This proves (b). Now $G_{\alpha}N$ is a Frobenius group with $G_{\alpha}$ as complement, and, if $|\Sigma| = p$, it is normal and its orbits are the orbits of $N$. This proves (c). Finally, if $N$ is not regular on $\Delta$ then it acts as a Frobenius group on $\Delta$ and thus on all its orbits. By the uniqueness of the Frobenius representation if follows that there are only $p$ orbits. This proves the first part of (d). The second part holds because no element of $G$ outside $N$ fixes more than $p$ orbits of $N$. $\square$

I have some difficulties, I highlighted the part's I do not understand fully. First in the proof of (a), why is $\overline a$ a minimal block, I know it is a block, but why minimal? The rest of the proof for part (a) I have understood I think. But then in the proof of part (b), I do not understand how exactly the lemma enters here and implies that $|\Delta'|$ and $|G_{\alpha}|$ are coprime. And why does Lagrange's theorem implies that the stabilizer is $G_{\alpha}N$, I see that by $$(\alpha^N)^{G_{\alpha}N} = \alpha^{G_{\alpha}NN} = \alpha^{NN} = \alpha^N$$ the subgroup $G_{\alpha}N$ is contained in the stabilizer, but I do not see equality? And what has this to do about the number of fixed orbits? For parts (c) and (d) I understand almost nothing, why does $|\Sigma| = p$ implies the assertions?

Would be glad if someone can explain these parts in more detail to me, so I can understand them?

• $\overline{\alpha}$ a minimal block because it has prime order $p$. The lemma says that $|\Omega|$ is congruent to $p$ mod $|G_\alpha|$, and $|\Delta'|$ divides $|\Omega|$, but $p$ does not divide $|\Delta'|$, so $|\Delta'|$ and $|G_\alpha|$ must be coprime. Sorry, that's all I have time for right now! – Derek Holt Oct 19 '15 at 19:06
• I would be glad if you find time to look at the rest! Regarding your comment, it might be simple number theory but $|\Omega|$ is congruent to $p$ mod $|G_{\alpha}|$ means $|G_{\alpha}|$ divides $|\Omega| - p$, i.e. $k |G_{\alpha}| = |\Omega| - p$, also we have $|\Omega| = l\cdot |\Delta'|$, so that $k|G_{\alpha}| = |\Delta'|\cdot l - p$. But why does this excludes that $|G_{\alpha}|$ and $|\Delta'|$ have a common divisor $> 1$? – StefanH Oct 19 '15 at 19:23
• Yes it's very simple number theory! For your next point, anything in the stabilizer of $\Delta$ maps $\alpha$ to a point in $\alpha^N$, so it lies in $G_\alpha N$. – Derek Holt Oct 19 '15 at 20:33
• Yes, now I see it... if $|G_{\alpha}|$ and $|\Delta'|$ would have a common divisor (which could not be $p$ by presupposition), this one would also divide $p$, hence it must be one. Thanks for your other hint, but do you know why they refer to Lagrange's theorem, as then this isn't used there if I see it correctly? – StefanH Oct 19 '15 at 20:52