[proofplan] The proof verifies the five assertions in turn. Statements (1), (3), and (4) follow directly from the definition of the lift $\rho := \tilde{\rho} \circ \pi$, where $\pi: G \twoheadrightarrow G/N$ is the quotient. The irreducibility equivalence in (2) is the crux: we compute $\langle \chi, \chi \rangle$ on $G$ by grouping the sum over $g \in G$ into cosets of $N$. Since $\chi$ is constant on cosets of $N$ (the lift factors through the quotient), each coset contributes $|N|$ identical terms, and the result simplifies to $\langle \tilde{\chi}, \tilde{\chi} \rangle$ on $G/N$. The bijection in (5) follows by noting that the lift always satisfies $N \leq \ker \chi$, and conversely any irreducible character with $N \leq \ker \chi$ descends to $G/N$ via the universal property of the quotient. [/proofplan]

[step:Define the lift and verify it is a representation of $G$]Let $\pi: G \to G/N$ denote the canonical quotient homomorphism, $g \mapsto gN$, which is well-defined because $N \lhd G$. Define the **lift** \begin{align*} \rho: G &\to \operatorname{GL}(V), \\ g &\mapsto \tilde{\rho}(gN) = (\tilde{\rho} \circ \pi)(g). \end{align*} Since $\pi: G \to G/N$ is a group homomorphism and $\tilde{\rho}: G/N \to \operatorname{GL}(V)$ is a group homomorphism, the composition $\rho = \tilde{\rho} \circ \pi$ is a group homomorphism $G \to \operatorname{GL}(V)$, hence a representation of $G$. This proves (1). For (3): for every $g \in G$, \begin{align*} \chi(g) = \operatorname{tr}(\rho(g)) = \operatorname{tr}(\tilde{\rho}(gN)) = \tilde{\chi}(gN), \end{align*} which is the formula in the statement. For (4): apply (3) at $g = 1_G$: \begin{align*} \deg \chi = \chi(1_G) = \tilde{\chi}(1_G \cdot N) = \tilde{\chi}(1_{G/N}) = \deg \tilde{\chi}, \end{align*} since $1_G \cdot N = N$ is the identity coset, and the degree of any character is its value at the group identity.[/step]

[guided]The lift $\rho$ is the composition $\tilde{\rho} \circ \pi$. The composition of two group homomorphisms is a group homomorphism, hence a representation of $G$ on the same vector space $V$ on which $\tilde{\rho}$ acts. This is (1). For the character formula (3), the key is that $\rho(g)$ is by construction equal to $\tilde{\rho}(gN)$ — the very same operator. So tracing both sides: \begin{align*} \chi(g) = \operatorname{tr}(\rho(g)) = \operatorname{tr}(\tilde{\rho}(gN)) = \tilde{\chi}(gN). \end{align*} The character of the lift at $g$ is the character of the original at the coset $gN$. For (4), the degree is the value of the character at the identity. Setting $g = 1_G$ in (3): $\chi(1_G) = \tilde{\chi}(1_G \cdot N) = \tilde{\chi}(N) = \deg \tilde{\chi}$, since $1_G \cdot N = N$ is the identity element of $G/N$. Equivalently, $\rho$ and $\tilde{\rho}$ map into the same $\operatorname{GL}(V)$, hence have the same degree $\dim V$.[/guided]

[step:Group $\langle \chi, \chi \rangle$ over cosets of $N$ to reduce to $\langle \tilde{\chi}, \tilde{\chi} \rangle$]Each $g \in G$ lies in a unique coset $gN \in G/N$, and the cosets partition $G$: \begin{align*} G = \bigsqcup_{xN \in G/N} xN. \end{align*} Each coset $xN$ has cardinality $|N|$. So \begin{align*} |G| = |G/N| \cdot |N|. \end{align*} For any $g \in G$ and $k \in N$, by (3), \begin{align*} \chi(gk) = \tilde{\chi}((gk)N) = \tilde{\chi}(gN \cdot kN) = \tilde{\chi}(gN \cdot N) = \tilde{\chi}(gN) = \chi(g), \end{align*} using $kN = N$ for $k \in N$ (the identity coset), and $gN \cdot N = gN$ in $G/N$. Hence $\chi$ is **constant on cosets of $N$**. Now compute, grouping the sum $\sum_{g \in G}$ by cosets: \begin{align*} \langle \chi, \chi \rangle &= \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)}\, \chi(g) \\ &= \frac{1}{|G|} \sum_{xN \in G/N} \sum_{k \in N} \overline{\chi(xk)}\, \chi(xk) \\ &= \frac{1}{|G|} \sum_{xN \in G/N} \sum_{k \in N} \overline{\tilde{\chi}(xN)}\, \tilde{\chi}(xN) \\ &= \frac{1}{|G|} \sum_{xN \in G/N} |N|\, |\tilde{\chi}(xN)|^2. \end{align*} Substituting $|G|/|N| = |G/N|$: \begin{align*} \langle \chi, \chi \rangle = \frac{|N|}{|G|} \sum_{xN \in G/N} |\tilde{\chi}(xN)|^2 = \frac{1}{|G/N|} \sum_{xN \in G/N} |\tilde{\chi}(xN)|^2 = \langle \tilde{\chi}, \tilde{\chi} \rangle, \end{align*} where the last equality is the definition of the inner product on class functions of $G/N$.[/step]

[guided]The strategy for (2) is to compare the inner product computed on $G$ with the inner product computed on $G/N$. Because $\chi$ is the lift of $\tilde{\chi}$, $\chi$ takes only $|G/N|$ distinct values, each repeated $|N|$ times (once per element of the coset). When we compute $\langle \chi, \chi \rangle = \frac{1}{|G|} \sum_g |\chi(g)|^2$, this redundancy collapses. Step (i): observe that $\chi$ is constant on cosets of $N$. For $g \in G$ and $k \in N$, $\chi(gk) = \tilde{\chi}((gk)N) = \tilde{\chi}(gN)$ since $kN = N$ in $G/N$. So all $|N|$ elements of any coset $xN$ give the same value $\chi = \tilde{\chi}(xN)$. Step (ii): partition $G$ by cosets and compute. The sum $\sum_{g \in G}$ becomes a double sum: outer over coset representatives $xN$, inner over the $|N|$ elements of each coset. The summand $|\chi(g)|^2 = |\tilde{\chi}(xN)|^2$ is constant in the inner sum, so the inner sum collapses to $|N| \cdot |\tilde{\chi}(xN)|^2$. The factor $|N|$ then combines with the prefactor $\frac{1}{|G|}$ to give $\frac{1}{|G/N|}$: \begin{align*} \langle \chi, \chi \rangle = \frac{|N|}{|G|} \sum_{xN \in G/N} |\tilde{\chi}(xN)|^2 = \frac{1}{|G/N|} \sum_{xN \in G/N} |\tilde{\chi}(xN)|^2 = \langle \tilde{\chi}, \tilde{\chi} \rangle. \end{align*} The result is the inner product of $\tilde{\chi}$ with itself on the quotient group.[/guided]

[step:Conclude the irreducibility equivalence]By [Row Orthogonality](/theorems/2430), an irreducible character $\psi$ of any finite group satisfies $\langle \psi, \psi \rangle = 1$. Conversely, by the [Irreducibility Criterion](/theorems/2424) (a consequence of completeness of irreducible characters), a character $\psi$ of a finite group is irreducible if and only if $\langle \psi, \psi \rangle = 1$. By Step 2, $\langle \chi, \chi \rangle = \langle \tilde{\chi}, \tilde{\chi} \rangle$. So \begin{align*} \chi \text{ is irreducible} \iff \langle \chi, \chi \rangle = 1 \iff \langle \tilde{\chi}, \tilde{\chi} \rangle = 1 \iff \tilde{\chi} \text{ is irreducible}, \end{align*} which is (2).[/step]

[guided]The irreducibility criterion is: $\psi$ is irreducible iff $\langle \psi, \psi \rangle = 1$. We have just shown $\langle \chi, \chi \rangle = \langle \tilde{\chi}, \tilde{\chi} \rangle$. So one is $1$ iff the other is, and the irreducibility statuses match. We could equivalently argue at the level of representations: invariant subspaces of $V$ under $\rho = \tilde{\rho} \circ \pi$ are the same as invariant subspaces under $\tilde{\rho}$, since $\rho(G) = \tilde{\rho}(\pi(G)) = \tilde{\rho}(G/N)$ (because $\pi$ is surjective). So irreducibility — meaning no proper non-zero invariant subspace — passes between $\rho$ and $\tilde{\rho}$ identically. The character calculation is cleaner because it avoids invoking surjectivity of $\pi$ directly.[/guided]

[step:Establish the bijection between characters of $G/N$ and characters of $G$ vanishing on $N$ in the kernel sense]For (5), define \begin{align*} L: \{\text{irreducible characters of } G/N\} &\to \{\text{irreducible characters of } G \text{ with } N \leq \ker \chi\}, \\ \tilde{\chi} &\mapsto \chi := \tilde{\chi} \circ \pi. \end{align*} **$L$ is well-defined.** By (2), if $\tilde{\chi}$ is an irreducible character of $G/N$ then its lift $\chi = L(\tilde{\chi})$ is an irreducible character of $G$. We must check $N \leq \ker \chi$, where $\ker \chi := \{g \in G : \chi(g) = \chi(1)\}$. For $k \in N$, $\chi(k) = \tilde{\chi}(kN) = \tilde{\chi}(N) = \tilde{\chi}(1_{G/N}) = \chi(1)$, so $k \in \ker \chi$. Hence $N \leq \ker \chi$. **$L$ is injective.** Suppose $L(\tilde{\chi}_1) = L(\tilde{\chi}_2)$. Then for every $g \in G$, $\tilde{\chi}_1(gN) = \tilde{\chi}_2(gN)$. Since $\pi: G \to G/N$ is surjective, every element of $G/N$ is of the form $gN$ for some $g \in G$, so $\tilde{\chi}_1 = \tilde{\chi}_2$ as functions on $G/N$. **$L$ is surjective.** Let $\chi$ be an irreducible character of $G$ afforded by an irreducible representation $\rho: G \to \operatorname{GL}(V)$, with $N \leq \ker \chi$. We claim $N \leq \ker \rho$, where $\ker \rho := \rho^{-1}(\{I\}) = \{g : \rho(g) = I\}$. [claim:If $\chi$ is the character of an irreducible representation $\rho$ of a finite group, then $\ker \chi = \ker \rho$] [proof] For any $g \in G$, $\rho(g)$ has finite order (since $G$ is finite), so its eigenvalues are roots of unity, of modulus $1$. The trace $\chi(g) = \sum_i \lambda_i$ is a sum of $\dim V$ unit complex numbers, and $\chi(1) = \dim V$. We have $|\chi(g)| = |\sum \lambda_i| \leq \sum |\lambda_i| = \dim V = \chi(1)$, with equality if and only if all $\lambda_i$ are equal positive real multiples of a common direction, i.e. all equal. Combined with $|\lambda_i| = 1$, equality requires all eigenvalues $\lambda_i = \zeta$ for a common root of unity $\zeta$. If $\chi(g) = \chi(1)$ (the condition for $g \in \ker \chi$), this is real and positive, so $\zeta = 1$ and all $\lambda_i = 1$. Hence $\rho(g) = I$ (a diagonalisable operator with all eigenvalues $1$ is the identity), i.e. $g \in \ker \rho$. Conversely, $g \in \ker \rho$ gives $\rho(g) = I$ directly, so $\chi(g) = \operatorname{tr}(I) = \dim V = \chi(1)$ and $g \in \ker \chi$. Therefore $\ker \chi = \ker \rho$. [/proof] [/claim] By the claim, $N \leq \ker \chi = \ker \rho$. Define \begin{align*} \tilde{\rho}: G/N &\to \operatorname{GL}(V), \\ gN &\mapsto \rho(g). \end{align*} Well-definedness: if $gN = g'N$, then $g^{-1}g' \in N \leq \ker \rho$, so $\rho(g^{-1}g') = I$, hence $\rho(g) = \rho(g')$, so $\tilde{\rho}(gN) = \tilde{\rho}(g'N)$. Homomorphism: $\tilde{\rho}((gN)(g'N)) = \tilde{\rho}(gg'N) = \rho(gg') = \rho(g)\rho(g') = \tilde{\rho}(gN)\tilde{\rho}(g'N)$. Hence $\tilde{\rho}$ is a representation of $G/N$ on $V$. The lift of $\tilde{\rho}$ is, by construction, $\rho$ itself. So the character $\tilde{\chi}$ of $\tilde{\rho}$ satisfies $L(\tilde{\chi}) = \chi$. Moreover, the irreducibility of $\rho$ — i.e. of $\chi$ — pulls back to irreducibility of $\tilde{\rho}$ by (2) (or equivalently because $\tilde{\rho}$ has the same image as $\rho$, hence the same invariant subspaces). So $\tilde{\chi}$ is an irreducible character of $G/N$. This shows every irreducible character of $G$ with $N \leq \ker \chi$ is of the form $L(\tilde{\chi})$, completing surjectivity. Hence $L$ is a bijection.[/step]

[guided]For (5), we need a bijection between two sets: \begin{align*} A &= \{\text{irreducible characters of } G/N\}, \\ B &= \{\text{irreducible characters } \chi \text{ of } G \text{ with } N \leq \ker \chi\}. \end{align*} The forward map $L: A \to B$ is the lift, $L(\tilde{\chi})(g) := \tilde{\chi}(gN)$. We need to verify three things: $L$ takes values in $B$, $L$ is injective, $L$ is surjective. For $L(A) \subseteq B$: by (2), the lift of an irreducible is irreducible. The condition $N \leq \ker \chi$ is the assertion that $\chi(k) = \chi(1)$ for all $k \in N$. By the lift formula, $\chi(k) = \tilde{\chi}(kN) = \tilde{\chi}(N) = \tilde{\chi}(1_{G/N}) = \chi(1)$. So $L(\tilde{\chi}) \in B$. For injectivity: $L$ is induced by composition with the surjection $\pi$. Any function $\tilde{f}: G/N \to \mathbb{C}$ is determined by its composition with $\pi$ (since $\pi$ is surjective). So $L(\tilde{\chi}_1) = L(\tilde{\chi}_2)$ implies $\tilde{\chi}_1 \circ \pi = \tilde{\chi}_2 \circ \pi$, hence $\tilde{\chi}_1 = \tilde{\chi}_2$. For surjectivity: given $\chi \in B$ afforded by an irreducible $\rho: G \to \operatorname{GL}(V)$, the condition $N \leq \ker \chi$ is equivalent (and this is the substantive step) to $N \leq \ker \rho$. The kernel of a character (defined by $\chi(g) = \chi(1)$) equals the kernel of the representation. This is a standard fact about complex characters of finite groups, established in the [claim] inside the step: a complex root of unity sum equals its maximum modulus only when all summands are equal, forcing $\rho(g) = I$. Once we have $N \leq \ker \rho$, the universal property of the quotient gives a unique homomorphism $\tilde{\rho}: G/N \to \operatorname{GL}(V)$ with $\tilde{\rho} \circ \pi = \rho$. By construction, $L(\tilde{\chi}) = \chi$ where $\tilde{\chi}$ is the character of $\tilde{\rho}$, and irreducibility of $\rho$ forces irreducibility of $\tilde{\rho}$ via (2). So $\chi$ is in the image of $L$. The bijection $L$ is the precise sense in which "characters of $G$ whose kernel contains $N$" coincides with "characters of $G/N$".[/guided]