On Alperin's paper "The Green Correspondence and Brauer's Characterization of Characters" (aka what is a central factor?) I was studying the paper "The Green Correspondence and Brauer's Characterization of Characters" by J. Alperin and I couldn't understand two of the passages.
Hypotheses and notations
$G$ is a finite group with the following properties:

*

*$G$ is not nilpotent;

*Every proper homomorphic image of $G$ is elementary;

*$G$ has a unique minimal normal subgroup $N$;

*$N$ is a $p$-group for some prime number $p$ (and thus it's elementary abelian).

First unclear passage
At this moment of the paper, Alperin is proving that $N = O_p(G)$, the $p$-core of $G$ (that is, the largest normal $p$-subgroup of $G$). To do so, he argues that $O_p(G)$ is an elementary abelian group in the following fashion:
"If $O_p(G)$ is not elementary abelian then the Frattini subgroup $D(O_p(G))$ of $O_p(G)$ is not $1$ so must contain $N$, by (3). But then $O_p(G)/D(O_p(G))$ is a central factor of $G$ so that $N$ will be also. Hence, $N$ is a central factor and $G/N$ is nilpotent so we have contradicted (1)."
The terminology "central factor" was new to me and I had to search for possible definitions. The only one that made sense in this context was the following: if $H,K \trianglelefteq G$ are normal subgroups and $K \leq H$, we say that $H/K$ is a central factor of $G$ if $H/K \leq Z(G/K)$. Based on this definition, I assumed that a (normal) subgroup of $G$ is a central factor if it's contained in $Z(G)$ (this agrees with the definition of central factor in Groupprops Subwiki if the normal subgroup is abelian, which is our case). Is there any other definition which makes more sense here?
I can see that $D(O_p(G))$ is not trivial and contains $N$. I also agree that $O_p(G)/D(O_p(G))$ is a central factor of $G$, since $G/D(O_p(G))$ is nilpotent by (2) so $O_p(G)/D(O_p(G))$ is the $p$-Sylow subgroup of $G/D(O_p(G))$ and is abelian. But how this implies that $N$ is a central factor? (Knowing this, I am able to understand the final sentence of the argument.)
Second unclear passage
The paper continues and Alperin has just proved that $N = O_p(G)$. He then invokes the Schur-Zassenhaus theorem and concludes that $G$ is the semi-direct product of $N$ and another subgroup $K$ (but I think this is irrelevant for my question). We arrive at this sentence:
"The uniqueness of $N$ now gives us that $N$ is not a central factor of $G$; hence, no non-trivial character of $N$ is stabilized by $G$."
Since $Z(G)$ is normal in $G$, it must contain $N$ if it isn't trivial. But then we could argue that $G$ is nilpotent as before, a contradiction. Thus, $Z(G) = 1$ and $N$ is clearly not a central factor. This intuitively implies that no non-trivial irreducible character of $N$ is stabilized by $G$, but I cannot give a proof. Could someone help me here too?
I said that it was "intuitive" so here is what my intuition says: We know that $N$ is elementary abelian so we understand what the irreducible characters of $N$ look like. If $\chi$ is a non-trivial character of $N$, there is $n \in N$ such that $\chi(n) \neq 1$ and hence $\chi(n)$ equals some primitive $p$-th root of unity. Since $G$ has trivial center, there is $g \in G$ such that $gng^{-1} \neq n$. It is reasonable to expect that $\chi(gng^{-1}) \neq \chi(n)$ (which would imply that $g$ does not stabilize $\chi$) but $\chi$ can indeed assume repeated values and this inequality is not (immediately) guaranteed. We would need to search for the right $n$ and the right $g$. Instead of dealing with individual elements of $N$, I also tried working with the isomorphism between $N$ and its dual group. If this isomorphism were compatible with the actions of $G$, the result would follow from the fact that $Z(G) = 1$. But this isomorphism is not canonical, so you can correctly guess that my calculations didn't take me where I wanted.
Maybe there is some combinatorial argument or maybe we have enough information to understand the action of $G$ on $N$, I don't know...
To sum it up

*

*Which definition of "central factor" is Alperin probably using?

If I am using the correct one:


*How to prove that $N$ is a central factor in the first passage?

*How to guarantee that the non-trivial irreducible characters of $N$ aren't fixed by $G$ in the second passage?

 A: After David A. Craven's very helpful commentaries and after some more research, I can now understand the unclear passages of the article, as I will explain below.
Definition of central factor
It seems that Alperin is probably using the definition of central factor I gave, since both passages can be proved with it.
First unclear passage
(I'll just expand on what David A. Craven wrote)
We know that $O_p(G)/D(O_p(G))$ is a central factor and want to prove that $N$ is also a central factor.
By the hypothesis, any element of $G$ centralizes $O_p(G)/D(O_p(G))$. Hence, by this theorem, any $p'$-element of $G$ centralizes $O_p(G)$. Remember that any element of $G$ can be written as a product of a $p'$-element with a $p$-element, which is an element of $O_p(G)$ (we already knew that $O_p(G)$ was the only $p$-Sylow of $G$). This easily implies that any normal subgroup of $O_p(G)$ is also normal in $G$, and we conclude that $N$ is the unique minimal normal subgroup of $O_p(G)$. But $Z(O_p(G))$ is non-trivial and normal in $O_p(G)$ so $N \subseteq Z(O_p(G))$. We now know that any element of $N$ commutes with any $p$-element and with any $p'$-element, implying that $N \subseteq Z(G)$, as desired.
Second unclear passage
My attempt of showing that $N$ and its dual group were isomorphic as $G$-sets (i.e., sets with an action by $G$) was in fact a fruitful direction to attack the problem. In a general context, these two $G$-sets need not to be isomorphic, but they will be in our case.
Since $N$ is abelian, it is in the kernel of the action by $G$ and we may regard $N$ as a $(G/N)$-set. At this point of the article, we know that $N$ is a $p$-Sylow of $G$, so $\textrm{gcd}(|N|,|G/N|) = 1$ and we can apply the following result:
Theorem 18.10 from B. Huppert's "Character Theory of Finite Groups": Let a group $A$ act on the abelian group $G$ (by automorphisms) and suppose $\textrm{gcd}(|G|,|A|) = 1$. Then $G$ and $\textrm{Irr}(G)$ are isomorphic $A$-sets.
(If $\chi \in \textrm{Irr}(G)$, an element $a \in A$ takes $\chi$ to the character $\chi^a$ given by $\chi^a(g) = \chi(g^{a^{-1}})$ for $g \in G$. The book considers right actions but everything can be transformed into left actions easily.)
In our case, the hypotheses above are satisfied if we change $A$ and $G$ by $G/N$ and $N$, respectively. Thus, $N$ and $\textrm{Irr}(N)$ are isomorphic as $(G/N)$-sets and so as $G$-sets too. But we know that $Z(G) = 1$, so $1$ is the only element of $N$ fixed by $G$. Consequently, $\textrm{Irr}(N)$ has just one element fixed by $G$, which must be the trivial character.
