0
$\begingroup$

I have just started learning about group representations, and I've come across some confusing language:

$\rho : G \rightarrow GL(V)$ factors via a representation of $G \big/ ker(\rho)$

I can see how this representation "induces" a representation on the quotient, is that what it means? The word factor makes me think you can write a product or decompose something (similar to Maschke's theorem).

Just for context $G = BD_{4m}$ where $m$ is even and $\rho$ maps into $GL(2, \mathbb{C})$ with kernel $\{\pm 1 \}$, so $G \big/ ker(\rho) \cong D_{2m}$

$\endgroup$
  • 1
    $\begingroup$ By factors, they mean that the map $\rho$ can be written as a composition of the projection from $G$ to $G/\operatorname{ker}(\rho)$ and a representation $\rho'$ of that quotient. So in this way, $\rho$ is a product of maps, i.e. it has been factored. $\endgroup$ – Tobias Kildetoft Jan 25 '17 at 11:57
  • $\begingroup$ I see, so does it have anything to do with irreducibility of the representation? $\endgroup$ – chilliBeanDream Jan 25 '17 at 12:20
  • 1
    $\begingroup$ No, this is completely unrelated to the reducibility of the representation. $\endgroup$ – Tobias Kildetoft Jan 25 '17 at 12:30
  • $\begingroup$ Thank you, you could put this as an answer and I'll accept it. $\endgroup$ – chilliBeanDream Jan 25 '17 at 18:07
0
$\begingroup$

This is just the statement of the first isomorphism theorem. Since $$\rho: G\to GL(V)$$ is a homomorphism, there exists a unique (injective) homomorphism $G/\ker\rho\to GL(V)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.