Let $H$ be a finite group, Let $V$ be a finite dimensional vector space over $\mathbb{C}$, and $\rho$ a representation of $H$ on $V$. Is there, for any representation $\rho$ of $H$ on $V$, another finite group $G$, $H < G$, such that $\tau$ is a represenation of $G$ on $V$, $\tau(h) = \rho(h) \ \forall h \in H$, and $\tau$ is irreducible?
My intuition says this is false, but I have had no traction whatsoever in proving it.
The question for general groups has been answered in the positive. If $H$ is a finite group and you insist that $G$ is also a finite group, however, then the answer is no, in general.
Let $H = 2.A_5 \times \mathbf{Z}/p \mathbf{Z}$ for any prime $p > 7$. (Here $2.A_5 = \mathrm{SL}_2(\mathbf{F}_5)$ is a non-trivial central extension of $A_5$.) The group $2.A_5$ has a faithful representation of dimension $2$, and $\mathbf{Z}/p \mathbf{Z}$ has a faithful representation of dimension $1$. Let $V$ be the direct sum of these representations, which is reducible of dimension $3$. Then $V$ as a representation of $H$ is not the restriction of any irreducible representation of a finite group $G$. We may assume that $G$ acts faithfully on $V$, since otherwise we can replace $G$ by the image, which still contains $H$ because $H$ acts faithfully. I claim that the action of $G$ on $V$ is primitive, that is, it is not induced from a subgroup $P$ with $[G:P] > 1$. Since $3 = \mathrm{dim}(V)$ is prime, $V$ would necessarily be induced from a character $\chi$ of an index $3$ subgroup $P$. But this would force $G$ to be solvable: it admits a map to $S_3$ (acting on the left cosets $G/P$), and the restriction of $V$ to the kernel $N$ is just a direct sum of characters, namely $\chi |_{N}$ and its conjugates by $G$. Since $N$ acts faithfully, this means that $N$ is abelian, and thus $G$ is solvable. Since $H$ is not solvable, this is impossible. On the other hand, there are only finitely many subgroups of $\mathrm{GL}_n(\mathbf{C})$ which act primitively. For $n = 3$, they have projective image one of six groups, the non-simple examples being $A_5$, $A_6$, and $\mathrm{PSL}_2(\mathbf{F}_7)$. (See finite subgroups of PGL(3,C)) Since the element of order $p$ in $H$ does not act diagonally on $V$, it acts non-trivially on the projective representation. For $p$ big enough ($p > 7$ in this case) this is a contradiction with the possible primitive subgroups of $\mathrm{PGL}_3(\mathbf{C})$.