Why are there $|G/G'|$ 1-dimensional representations of $G$? Let $G'$ be the derived subgroup of a finite group $G$. We have a correspondence $\\{\mathrm{reps \ of \ G/G'}\\} \longleftrightarrow \\{\mathrm{reps \ of \ G \ with \ kernel \ containing \ G' }\\} $ If we restrict to 1-dimensional reps, we get: $\\{\mathrm{1\ dimensional \ reps \ of \ G/G'}\\} \longleftrightarrow \\{\mathrm{1 \ dimensional \ reps \ of \ G \ with \ kernel \ containing \ G' }\\} $ Now my notes say that there are $|G/G'|$ 1-dimensional reps of $G$. Since there are $|G/G'|$ 1-dimensional reps of $G/G'$, this must mean that all 1-dimensional reps of $G$ have kernel containing $G'$. Why is this so? Thanks

The derived group $G^\prime$ is generated by the commutators, i.e. the elements of the form $ghg^{-1}h^{-1}$.

A 1-dimensional representation is a _character_ , i.e. an homomorphism $$ \rho:G\longrightarrow\Bbb C^\times. $$ Since $\Bbb C^\times$ is abelian, $G^\prime<\ker(\rho)$.

Also, $G/G^\prime$ is abelian, so the number of its characters coincides with the number of its elements.

Putting all things together, $G$ has $|G/G^\prime|$ one dimensional representations.