Let $G/H\cong PSL(2,11)$, and $\theta$ be an irreducible $\mathbb{C}$-character of $H$. Suppose $\theta$ is invariant in $G$ and $\theta(1)=9$.



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> Question: Is $\theta$ extendible to $G$?
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Is a $G$-invariant character $\theta$ of $H$ extendible to $G$?
The answer is yes (and some of the comments were moving in the right direction): Let $T$ be a transversal to $H$ in $G,$ and let $\sigma$ afford the representation of $H.$ For each $t \in T,$ there is a matrix $M\_{t} \in {\rm GL}(9,\mathbb{C})$ such that $\sigma(tht^{-1}) = M\_{t}\sigma(h)M\_{t}^{-1}$ for all $h \in H,$ and $M\_{t}$ is unique up to scalar multiples (by Schur's Lemma) , while any non-zero scalar multiple will have the same property. Extend this to $G$ by letting $M\_{th} = M\_{t}\sigma(h)$ for each $t \in T, h \in H.$
(It might be convenient here for anyone interested in full detail, to assume that have multiplied $\sigma$ by a suitable power of the linear character $ \lambda = {\rm det} \sigma,$ so that ${\rm det} \sigma$ may be assumed to have multiplicative order a power of $3$).


Note also that for $x,y \in G,$ there is a scalar $\alpha(x,y) \neq 0$ such that $M\_{xy} = \alpha(x,y)M\_{x}M\_{y} (\ast).$ 


Notice then $M\_{x}^{|G|}$ is a scalar matrix for each $x \in G.$ Multiplying each $M\_{x}$ by a suitable scalar (and we can still keep $M\_{h} = \sigma(h)$ for each $h \in H$, and $M\_{th} = M\_{t}\sigma(h)$ for $t \in T,h \in H$), we may, and do from now on, assume that each $M\_{x}$ has determinant a $3$-power root of unity. 


It follows from $(\ast)$ (on taking determinants), that $\alpha(x,y)$ is a $3$-power root of unity for all $x,y \in G.$ This gives a $2$-cocycle for ${\rm PSL}(2,11)$ of $3$-power order.


Now we can finish in either of two ways: the Schur multiplier of the perfect group ${\rm PSL}(2,11)$ is well-known to have order $2.$ But a more general argument is to note that a perfect group with a cyclic Sylow $p$-subgroup always has a Schur multiplier of order prime to $p$ ( and this may be applied here with $p =3$ ).


