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Let $G$ be a (finite) group and let $N$ be a normal subgroup of $G$.

Suppose that we have a representation $(V,\rho)$ of $N$ and a representation of $(V, \tau)$ of the quotient group $G/N$. Here $V$ is a vector space.

Question: Is there a way to construct a unique representation for $G$ that restricts to the representation $(V,\rho)$ of $N$?

In other words, I want to know if there is some equivalence (bijection?) between pairs of representations of $N$ and $G/N$ and a representation of $G$.

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    $\begingroup$ The question is a bit confusing in its current form. $\tau$ seems to be unconnected to the rest. If $\rho$ is the (trivial) representation of the trivial subgroup $N:=\{e\}$, then there are certainly more extensions to all of $G$, in general. $\endgroup$ Feb 11, 2016 at 23:39
  • $\begingroup$ @PeterFranek I edited the question. Does this make sense? $\endgroup$
    – user65175
    Feb 12, 2016 at 1:16

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There is a way to use information about $N$ and $G/N$ to study the representation theory of $G$, but it's less direct than this: the representation theory of $N$ is involved but $G/N$ is involved in a more complicated way. The keyword you want is Clifford theory.

If $N$ is abelian, the statement is the following. $G/N$ naturally acts on the set $\widehat{N}$ of irreducible (necessarily $1$-dimensional) complex representations of $N$, and the irreducible representations of $G$ are in natural bijection with the disjoint union of the irreducible representations of the stabilizers of the action of $G/N$ on $\widehat{N}$. This remains approximately true if $N$ is nonabelian, except that we need to consider projective representations.

The statement in the form you want it cannot be true for cardinality reasons. For example, if $G = S_3, N = C_3$, then there are $6$ pairs of an irreducible representation of $N$ and an irreducible representation of $G/N$, but $G$ itself only has $3$ irreducible representations. $G/N \cong C_2$ has two orbits acting on $\widehat{C_3}$, one of which has stabilizer $C_2$ and one of which has trivial stabilizer, so that $3$ is $2 + 1$.

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