Let \(H\trianglelefteq G\) with \(G\) finite. Lift every representation of \(G/H\) along the quotient map \(G\to G/H\).
- Show that the intersection of the kernels of all lifted representations is exactly \(H\).
- Interpret the result when \(H=[G,G]\).
Solution
Let \[ \pi:G\to G/H \] be the quotient map.
1. The common kernel is \(H\).
Every lifted representation has the form \[ \rho\circ\pi \] for some representation \(\rho\) of \(G/H\), so every such kernel contains \(H\).
Conversely, use the regular representation of \(G/H\). It is faithful: distinct elements of \(G/H\) act differently by left translation on the basis vectors of the group algebra. Hence the lifted regular representation has kernel exactly \[ \ker\pi=H. \] Therefore the intersection of the kernels of all lifted representations is exactly \(H\).
2. Take \(H=[G,G]\).
Then \[ G/H=G^{\mathrm{ab}} \] is the abelianization. Thus the representations of \(G\) factoring through the abelianization have common kernel exactly \[ [G,G]. \] Over \(\CC\), since a finite abelian group has only one-dimensional irreducible representations, this can be stated as \[ [G,G] = \bigcap_{\chi:G\to\CC^\times}\ker\chi, \] where the intersection runs over all complex linear characters of \(G\).