Let $U \le G$ be a subgroup of the finite group $G$ of odd order. And suppose $t \notin U$ is an involution normalising $U$, i.e. $U^t = U$ and $t^2 = 1$. Assume $t$ centralizes $U / U'$, i.e. $u^tU' = uU'$ for all $u \in U$ and let $\lambda \ne 1$ be a linear character of $U$. Then is it possible to extend it to a linear character of $TU$ (the subgroup which is formed as the semidirect product of $T$ and $U$) where $T = \langle t \rangle$?
Does the definition $\hat \lambda(tu) = \hat \lambda(u) = \lambda(u)$ for $u \in U$, i.e. it just depend on $u$, give a linear character of $TU$?
As pointed out by Derek in his comment, it is indeed not that difficult. Notice that for $u,v \in U$ we have $\lambda([u,v]) = [\lambda(u), \lambda(v)] = 1$, where $[u,v] = u^{-1}u^v$ is the commutator, and it equals the neutral element iff both arguments commute. This gives $\lambda(x) = 1$ for all $x \in U'$.
Further as $u^t \in uU'$ we have for $u, u' \in U$ that $ut = tux$ with $x \in U'$, hence $$ \hat \lambda(tutu') = \hat \lambda(ttuxu') = \hat \lambda(uxu') = \lambda(uxu') = \lambda(u)\lambda(x)\lambda(u') = \lambda(u)\lambda(u') = \hat \lambda(tu) \hat \lambda(tu'). $$ Also $$ \hat \lambda(tuu') = \lambda(uu') = \lambda(u) \lambda(u') = \hat \lambda(tu) \hat \lambda(u') $$ and $$ \hat \lambda(utu') = \hat \lambda(tuxu') = \lambda(uxu') = \lambda(u)\lambda(u') = \hat\lambda(tu)\hat\lambda(u') $$ which shows everything.