Exercise 6.17 from Isaac's Character Theory of Finite Groups. I am trying to solve the following problem in Isaac's Character Theory of Finite Groups and I am really sorry if this is a very simple question.
6.17: Let $N$ be a normal subgroup of $G$ with $G/N$ cyclic. Let $\eta\in \operatorname{Irr}(N)$ be invariant in $G$ and assume that $(\eta(1),|G:N|)=1.$ Show that $\eta$ is extendible to $G.$
Q1: The hint is to show that $G/\ker(\det\eta)$ is abelian. But I don't know how to prove the hint or how to solve the problem using it. I think proving that $\det \eta$ would be enough but I don't know how to do that.
Q2: There's a note right below the problem saying that this is true even without the assumption $(\eta(1),|G:N|)=1.$ Could someone please give me a hint for this? My only experience in Character Theory is reading up to chapter 6 in Isaac's book so I'm not sure if I know enough material to solve this without this assumption.
Thank you!!
 A: These are just hints.
To prove the hint in the book, note that $N/\ker(\det \eta) \le Z(G/\ker(\det \eta))$, since otherwise $\eta$ would not be invariant in $G$. Now $G/\ker(\det \eta)$ abelian follows from $G/N$ being cyclic.
To prove the result using the hint, use Theorem (6.25) in the book.
It is not part of the exercise to solve the problem without the assumption, so you should probably not worry about it! I have not thought about it myself, but it seems likely that the solution involves some results or techniques that have not been covered in the book at this point.
A: All credits go to prof Holt (+1 from me), but here is how I solved it a long time ago (hope it is still correct...). We need a lemma (actually more than we need, but this is not be found in Isaacs's book, I follow his notation and also the formulation of Problem (6.17) ...)
Lemma Let $N \unlhd G$ and $\vartheta \in Irr(N)$ be $G$-invariant. Put $\lambda=det(\vartheta)$. Then the following hold.
$(a)$ $\lambda$ is a $G$-invariant linear character of $N$
$(b)$ $ker(\vartheta) \subseteq ker(\lambda)$ and both subgroups are normal in $G$
$(c)$ $[G,N] \subseteq ker(\lambda)$ or equivalently, $N/ker(\lambda) \subseteq Z(G/ker(\lambda))$
$(d)$ $N/ker(\lambda)$ is cyclic and $|N:ker(\lambda)|=o(\lambda)$.
Proof (a) For all $g \in G, n \in N$, we have $\lambda^g(n)=\lambda(gng^{-1})=det(\vartheta)(gng^{-1})=det(\vartheta^g)(n)= \text { ($\vartheta$ is $G$-invariant) }det(\vartheta)(n) =\lambda(n)$.

(b) Let $\frak{T}$ be a representation of $N$ affording $\vartheta$ and let $n \in ker(\vartheta)$. Then $\frak{T}$$(n)=I$, and hence det($\frak{T}$$(n))=1$, so $ker(\vartheta) \subseteq ker(\lambda)$. For all $g \in G, n \in ker(\lambda)$, we have, using (a), $\lambda(gng^{-1})=\lambda^g(n)=\lambda(n)=1$, whence $gng^{-1} \in ker(\lambda)$. A similar argument shows that $ker(\vartheta) \unlhd G$.

(c) For all $g \in G, n \in N$ we have $\lambda([g,n])=\lambda(g^{-1}n^{-1}gn)=\lambda(g^{-1}n^{-1}g)\lambda(n)=\lambda^{g^{-1}}(n^{-1})\lambda(n)= \text { (by (a)) }\lambda(n^{-1})\lambda(n)=\lambda(1)=1.$

(d) Observe that since $\lambda$ is a linear character of $N$, $Z(\lambda)=N$. Now by Lemma(2.27)(d) (M.I. Isaacs, CTFG) $N/ker(\lambda)$ is cyclic. $\lambda$ is a faithful character of $N/ker(\lambda)$, so $o(\lambda)=|N:ker(\lambda)|. \square$
Now we are ready to prove the Problem(6.17): the Lemma(c) tells us that $N/ker(\lambda) \subseteq Z(G/ker(\lambda)) \subseteq G/ker(\lambda)$. But since $G/N$ is cyclic, it follows that $\overline{G}/Z(\overline{G})$ is cyclic (here $\overline{\cdot}$ denotes modding out by $ker(\lambda))$. This implies that $\overline{G}=G/ker(det(\vartheta))$ is abelian. Owing to Theorem (6.25) in Isaacs' book, the conclusion now follows, since $\overline{N}$ is a subgroup of the abelian group $\overline{G}$, and $\lambda$ can be viewed as a character of $\overline{N}$. (See also (5.5) Corollary in Isaacs' book)
The fact that the condition $gcd(\vartheta(1),|G:N|)=1$ can be dropped is explained in Chapter $11$, see Corollary (11.22). The theory here depends on projective representations, group extensions and hence a bit of cohomology theory (the so-called Schur multiplier $M(G)=H^2(G,\mathbb{C}^*)$ plays a role here, "measuring" certain extensions of the group $G$). 
The theory amounts to another powerful tool called character triples: an object $(G,N,\vartheta)$ is called a character triple, when $N \unlhd G$ and $\vartheta \in Irr(N)$ is $G$-invariant. Without going into the details (see Isaacs' CTFG page 186, and even better Isaacs' latest book Characters of Solvable Groups (CSG), where he, starting on p. 14, explains the essentials of the theory in 2.5 pages (!) (OK, for proofs he refers to his book CTFG)) one can define "isomorphisms" between character triples. A central theorem is then that every character triple $(G,N,\vartheta)$ admits an isomorphic one $(G^*,N^*,\vartheta^*)$, with $\vartheta^*$ faithful and linear and $N^* \subseteq Z(G^*)$. Here $G/N \cong G^*/N^*$. So the power of this tool lies in the fact that a certain assertion is made in the normal character world, then this is transposed to the $*$-world, where life is much easier so to speak, prove it there and cycle it back to the original assertion. The same tool allows you to drop the solvability condition in Problem (6.7), for example.
For further (historical) reading I suggest the aforementioned CSG and I.M. Isaacs, Characters of solvable and symplectic groups, Am. J. Math 95 (1973) 594-635, first couple of chapters. Here character triple isomorphisms are introduced. The rudiments of character triple theory was originally developed by Everett C. Dade if I am not mistaken.
