[proofplan] We verify the four assertions in turn. For (1), we show that $G'$ is closed under conjugation by computing $h[a,b]h^{-1}$ as a product of two commutators, which immediately lies in $G'$. For (2), the abelianness of $G/G'$ is exactly the assertion that all commutators $[g,h]$ lie in $G'$, which is the defining property of $G'$. For (3), we recognise that "$G/N$ is abelian" is equivalent to "$N$ contains every commutator", forcing $G' \leq N$. For (4), we apply the [Lifting Lemma](/theorems/2437) to transport the $\ell = |G:G'|$ irreducible characters of the abelian quotient $G/G'$ — all of degree one — to $\ell$ degree-one characters of $G$, and conversely show that every degree-one character of $G$ kills $G'$ and so descends to $G/G'$. [/proofplan]

[step:Show $G'$ is normal by writing conjugates of commutators as products of commutators] Recall that $G' = \langle [a,b] : a, b \in G \rangle$ is the subgroup generated by commutators $[a,b] := aba^{-1}b^{-1}$. To show $G' \lhd G$, it suffices to show $hG'h^{-1} \subseteq G'$ for all $h \in G$, and since $G'$ is generated by commutators it suffices to check that each conjugate $h[a,b]h^{-1}$ lies in $G'$. Compute, inserting $h^{-1}h = 1$ between strategic factors: \begin{align*} h[a,b]h^{-1} &= h\,a\,b\,a^{-1}\,b^{-1}\,h^{-1} \\ &= (h a)(b)(h a)^{-1}(h)(h)^{-1} \cdot h \, b^{-1} h^{-1} \\ &= [ha, b] \cdot h b^{-1} h^{-1}. \end{align*} The first factor $[ha, b] = (ha)b(ha)^{-1}b^{-1}$ is a commutator. For the second factor, write \begin{align*} h b^{-1} h^{-1} = (h b^{-1} h^{-1} b) \cdot b^{-1} = [h, b^{-1}] \cdot b^{-1}. \end{align*} Combining, \begin{align*} h[a,b]h^{-1} = [ha, b] \cdot [h, b^{-1}] \cdot b^{-1}. \end{align*} The right-hand side is not yet visibly in $G'$ because of the trailing $b^{-1}$. We instead use a cleaner computation: insert $b^{-1}b$ in the middle of $h[a,b]h^{-1}$. \begin{align*} h[a,b]h^{-1} &= (h a h^{-1})(h b h^{-1})(h a^{-1} h^{-1})(h b^{-1} h^{-1}) \\ &= [hah^{-1}, hbh^{-1}], \end{align*} since $(hah^{-1})(hbh^{-1})(hah^{-1})^{-1}(hbh^{-1})^{-1} = h\,aba^{-1}b^{-1}\,h^{-1}$ by direct expansion. Thus $h[a,b]h^{-1}$ is itself a commutator, hence in $G'$. Since the set of commutators is closed under conjugation, the subgroup $G'$ they generate is also closed under conjugation: for $w = c_1 c_2 \cdots c_m \in G'$ a word in commutators (and their inverses, but $[a,b]^{-1} = [b,a]$ is a commutator), $hwh^{-1} = (hc_1 h^{-1})(hc_2 h^{-1}) \cdots (h c_m h^{-1})$, each factor a commutator, so $hwh^{-1} \in G'$. Therefore $G' \lhd G$. [/step]

[step:Show $G/G'$ is abelian by reading off commutators in the quotient] For any $a, b \in G$, the commutator $[a,b] = aba^{-1}b^{-1}$ lies in $G'$ by definition. Reducing modulo $G'$: \begin{align*} [aG', bG'] = (aG')(bG')(aG')^{-1}(bG')^{-1} = (aba^{-1}b^{-1})G' = [a,b] G' = G', \end{align*} which is the identity in $G/G'$. Hence every commutator in $G/G'$ equals the identity coset, so $(aG')(bG') = (bG')(aG')$ for all $a,b$, i.e., $G/G'$ is abelian. [/step]

[step:Verify the universal property: $G/N$ abelian forces $G' \leq N$]Let $N \lhd G$. The quotient $G/N$ is abelian if and only if $(gN)(hN) = (hN)(gN)$ for all $g, h \in G$, which is equivalent to $ghg^{-1}h^{-1} N = N$, i.e., $[g,h] \in N$ for all $g, h \in G$. Since $G'$ is generated by all commutators, this last condition is equivalent to $G' \leq N$. In particular, taking $N = G'$ recovers (2), and the implication "$G/N$ abelian $\implies G' \leq N$" gives (3).[/step]

[guided]The universal property reads: among all normal subgroups $N$ of $G$ with abelian quotient, $G'$ is the smallest. The proof reduces to a single equivalence: \begin{align*} G/N \text{ abelian} \iff [g,h] \in N \text{ for all } g,h \in G \iff G' \subseteq N. \end{align*} The first equivalence is by definition of abelian: two cosets commute iff their commutator is the identity coset, i.e., the underlying commutator lies in $N$. The second equivalence uses that $G'$ is generated by commutators: a subgroup contains all commutators iff it contains the subgroup they generate. This is exactly the universal property of the abelianisation $G^{\mathrm{ab}} := G/G'$: any homomorphism from $G$ to an abelian group factors uniquely through $G^{\mathrm{ab}}$.[/guided]

[step:Count linear characters of $G$ via the Lifting Lemma applied to $G/G'$]Set $\ell := |G : G'|$. By (2), $G/G'$ is abelian of order $\ell$. A finite abelian group of order $\ell$ has exactly $\ell$ irreducible characters, all of degree $1$ — this follows from the [Number of Irreducible Characters](/theorems/2428) (the count equals the number of conjugacy classes, which for an abelian group is the order) combined with the constraint $\sum_i \chi_i(1)^2 = \ell$, forcing each $\chi_i(1) = 1$. By the [Lifting Lemma](/theorems/2437), the operation $\tilde{\chi} \mapsto \chi$ defined by $\chi(g) := \tilde{\chi}(gG')$ is a bijection \begin{align*} \{\text{irreducible characters of } G/G'\} &\longleftrightarrow \{\text{irreducible characters of } G \text{ with } G' \leq \ker \chi\}. \end{align*} The lift preserves degrees: $\deg \chi = \deg \tilde{\chi} = 1$. So the right-hand side contains exactly $\ell$ characters, all of degree $1$. It remains to show that every degree-one irreducible character of $G$ appears on the right-hand side, i.e., satisfies $G' \leq \ker \chi$. Let $\chi: G \to \mathbb{C}^\times$ be a degree-one irreducible character. Since $\chi$ is itself the (one-dimensional) representation, $\chi$ is a group homomorphism into the abelian group $\mathbb{C}^\times$. For any commutator $[a,b]$, \begin{align*} \chi([a,b]) = \chi(a)\chi(b)\chi(a)^{-1}\chi(b)^{-1} = 1, \end{align*} since $\mathbb{C}^\times$ is abelian. Hence $[a,b] \in \ker \chi$ for all $a,b$, and since $G'$ is generated by commutators, $G' \leq \ker \chi$. Combining: the set of degree-one irreducible characters of $G$ coincides with the image of the lift map, which has cardinality $\ell$. Therefore $G$ has exactly $\ell = |G:G'|$ irreducible representations of dimension $1$, namely the lifts of the irreducible characters of $G/G'$. By Lagrange's theorem, $\ell = |G|/|G'|$ divides $|G|$.[/step]

[guided]The argument has three pieces: (i) $G/G'$ is abelian and its character theory is elementary; (ii) the [Lifting Lemma](/theorems/2437) transports irreducible characters of the quotient to irreducible characters of $G$ that vanish on $G'$ (in the kernel sense); (iii) every linear character of $G$ vanishes on $G'$ in the kernel sense. For (i): an abelian group $A$ of order $\ell$ has $\ell$ conjugacy classes (each element is its own class), so by the [Number of Irreducible Characters](/theorems/2428) it has $\ell$ irreducible characters. The dimension formula $\sum_i \chi_i(1)^2 = |A| = \ell$ then forces each $\chi_i(1) = 1$, since the sum of $\ell$ positive squares equalling $\ell$ requires each to be $1$. For (ii): the [Lifting Lemma](/theorems/2437) states that lifts give a bijection $\{\text{irrs of } G/N\} \leftrightarrow \{\text{irrs of } G \text{ with } N \leq \ker\}$ preserving degree. Apply this with $N = G'$, $\ell$ irreducibles of degree $1$ on the left, and we get $\ell$ irreducibles of degree $1$ on the right. For (iii): the key observation is that any degree-one character $\chi: G \to \mathbb{C}^\times$ is itself a homomorphism to an abelian group. The universal property of $G/G'$ (Step 3) says any such homomorphism factors through $G/G'$, equivalently $G' \leq \ker \chi$. Concretely: \begin{align*} \chi([a,b]) = \chi(a)\chi(b)\chi(a)^{-1}\chi(b)^{-1} = 1 \end{align*} because the right-hand side is computed in $\mathbb{C}^\times$, which is abelian. Hence every commutator lies in $\ker \chi$, and so does the generated subgroup $G'$. The conclusion is that the lift bijection has image equal to the set of all linear characters of $G$. Counting: there are exactly $|G:G'|$ of them, and this number divides $|G|$ by Lagrange.[/guided]