Let $G$ be a finite group and $G'$ be the derived subgroup of $G.$ Then I know that every degree $1$ representation of $G$ factors through $G/G'.$ How does it imply that the number of degree $1$ representations of $G$ is same as $\left \lvert G/G' \right \rvert\ $?
Would anybody please help me in this regard? Thanks in advance.
A general fact for finite groups is that they have a finite number of irreduicible representations and if the dimensions of these are $d_1,..., d_n$ then we have $$d^2_1 + ... + d^2_n = |G|.$$ This is a consequence of Schur orthogonality relations among characters.
If $G$ is abelian, then all the irreducible reps are one dimensional and so the previous equation tells you $|G| = n$, the number of irreducible reps.
Now, in your context, $G/G'$ is abelian and as you say every 1 dimensional rep of $G$ corresponds, bijectively since you can always compose with the quotient, with those 1 -dimensional reps of $G/G'$, but his being abelian has $|G/G'|$ of those, by the argument above.