M.Isaacs Character Theory of Finite Groups Problem 6.18 Proof for Extendible to G Let $N\vartriangleleft G$ and suppose that $G=NH$ with  $N \cap H =1$.
Let  $\theta \in Irr(N)$ be invariant in $G$ and assume that  $(\theta(1), |G:N|)=1$.
If $H$ is solvable, show that  $\theta$ is extendible to $G$.
We know that $H\cong G/N$ solvable so by Theorem 6.25(Isaacs-Character Theory) $\theta$ is extendible to G iff $\det\theta$ is extendible to G. Hence how I show that $\det\theta$ is extendible to G?
OR
By Corollary 6.28  If I show that $(|G:N|, o(\theta)\theta(1))=1$, $\theta$ is extendible to G. But I don't know $(|G:N|, o(\theta))=1$ ?
Thank you in advance for your interest.
 A: Lemma Let $G=HN$, $H \leq G$, $N \unlhd G$, and $H \cap N=1$. Let $\mu \in Irr(N)$ be linear and invariant in $G$. Then $\mu$ can be extended to $G$.
Proof Since $H \cap N=1$, every $g \in G$ can be written as $g=hn$, with $h \in H, n\in N$ unique. Hence, define for $h \in H$, $n \in N$ $\lambda(hn)=\mu(n)$. Then $\lambda_N=\mu$. If $h_1,h_2 \in H$ and $n_1,n_2 \in N$, then $\lambda(h_1n_1h_2n_2)=\lambda(h_1h_2n_1^{h_2}n_2)=\mu(n_1^{h_2}n_2)=\mu(n_1^{h_2})\mu(n_2)= \text{ ($\mu$ is $G$-invariant) } \mu(n_1)\mu(n_2)=\lambda(h_1n_1)\lambda(h_2n_2).$
The Lemma has the following consequence.
(6.18)Exercise Let $N \unlhd G$ and suppose $G=NH$ with $N \cap H=1$. Let $\vartheta \in Irr(N)$ be $G$-invariant and assume $gcd(\vartheta(1),|G:N|)=1$. If $H$ is solvable, then $\vartheta$ is extendible to $G$.
Proof Since $G/N \cong H$ is solvable we can apply Theorem(6.25): we need to show that $det(\vartheta)$ is extendible. Since $\vartheta$ is $G$-invariant, so is $det(\vartheta)$, and the Lemma applies.
