[proofplan] The forward direction is immediate from the definition: isomorphic representations have the same trace at each $g$, hence the same character. For the converse, given $\rho$ with character $\chi$, we extract from $\chi$ enough data to recover $\rho$ up to isomorphism. The mechanism is: by [Maschke's Theorem](/theorems/2409), $\rho$ decomposes as a direct sum of irreducibles, $\rho \cong \bigoplus_j m_j \rho_j$, and the character decomposes correspondingly as $\chi = \sum_j m_j \chi_j$. The [Multiplicity Formula](/theorems/2420) (or equivalently row orthogonality) reads off the multiplicity $m_j$ as $\langle \chi, \chi_j \rangle$. Since the $\chi_j$ are an orthonormal family in $\mathcal{C}(G)$ (Row Orthogonality from [Completeness of Characters](/theorems/2424)), each $m_j$ is uniquely determined by $\chi$. The decomposition is unique up to permutation of summands, so $\rho$ is determined up to isomorphism. [/proofplan] [step:Establish the forward implication: isomorphic representations have equal characters] Suppose $\rho: G \to \operatorname{GL}(V)$ and $\rho': G \to \operatorname{GL}(V')$ are isomorphic complex representations of $G$. By definition, there exists a $\mathbb{C}$-linear isomorphism $\varphi: V \to V'$ such that \begin{align*} \varphi \circ \rho(g) = \rho'(g) \circ \varphi \quad \text{for every } g \in G. \end{align*} Equivalently, $\rho'(g) = \varphi \circ \rho(g) \circ \varphi^{-1}$. Taking traces and using conjugation-invariance of the trace ($\operatorname{tr}(ABA^{-1}) = \operatorname{tr}(B)$): \begin{align*} \chi'(g) = \operatorname{tr}(\rho'(g)) = \operatorname{tr}(\varphi \circ \rho(g) \circ \varphi^{-1}) = \operatorname{tr}(\rho(g)) = \chi(g). \end{align*} This holds for every $g \in G$, so $\chi = \chi'$ as functions on $G$. [/step] [step:Decompose $\rho$ into irreducibles via Maschke and read off the character] For the converse, fix a complex representation $\rho: G \to \operatorname{GL}(V)$ with character $\chi$, and let $\rho': G \to \operatorname{GL}(V')$ be a second complex representation with the same character $\chi' = \chi$. We show $\rho \cong \rho'$. Apply [Maschke's Theorem](/theorems/2409) to $\rho$: since $G$ is finite and $\operatorname{char} \mathbb{C} = 0$ does not divide $|G|$, the representation $\rho$ is completely reducible. Let $\rho_1, \ldots, \rho_k$ enumerate (up to isomorphism) the distinct irreducible complex representations of $G$, with characters $\chi_1, \ldots, \chi_k$ and underlying spaces $V_1, \ldots, V_k$. Then there exist non-negative integers $m_1, \ldots, m_k \geq 0$ (the multiplicities) and a $G$-equivariant isomorphism \begin{align*} V \cong \bigoplus_{j=1}^k V_j^{\oplus m_j} \quad \text{as } G\text{-representations}. \end{align*} Equivalently, $\rho \cong \bigoplus_{j=1}^k m_j \rho_j$, where $m_j \rho_j$ denotes the direct sum of $m_j$ copies of $\rho_j$. By Property 4 of [Elementary Properties of Characters](/theorems/2421) (additivity of characters under direct sum), iterated: \begin{align*} \chi = \chi_{\oplus_j m_j \rho_j} = \sum_{j=1}^k m_j\, \chi_j. \end{align*} Apply the same argument to $\rho'$, obtaining non-negative integers $m'_1, \ldots, m'_k \geq 0$ with $\rho' \cong \bigoplus_j m'_j \rho_j$ and $\chi' = \sum_j m'_j \chi_j$. [guided] The strategy is to express both $\rho$ and $\rho'$ in a canonical form — a direct sum of irreducibles indexed by their isomorphism class — and show that the character determines the multiplicities in this canonical form. Maschke's theorem is the engine: it guarantees complete reducibility for finite $G$ over $\mathbb{C}$. We verify Maschke's hypotheses: $G$ is finite (global hypothesis), and the field $\mathbb{C}$ has characteristic $0$, so $\operatorname{char}\mathbb{C} = 0 \nmid |G|$ for any $|G| \geq 1$. Hence Maschke applies to both $\rho$ and $\rho'$, and each decomposes as a direct sum of irreducibles. The character of a direct sum is the sum of characters (Property 4 of Elementary Properties of Characters). So \begin{align*} \chi = \sum_j m_j \chi_j, \qquad \chi' = \sum_j m'_j \chi_j, \end{align*} where $m_j$ and $m'_j$ are the multiplicities of $\rho_j$ in $\rho$ and $\rho'$ respectively. The hypothesis $\chi = \chi'$ gives $\sum_j m_j \chi_j = \sum_j m'_j \chi_j$. To extract $m_j = m'_j$ we need linear independence of the $\chi_j$ — which is row orthogonality. [/guided] [/step] [step:Apply Row Orthogonality to extract the multiplicities $m_j$] By the Row Orthogonality part (Part 1) of [Completeness of Characters](/theorems/2424), the irreducible characters $\chi_1, \ldots, \chi_k$ are pairwise orthogonal of unit norm in $\mathcal{C}(G)$: \begin{align*} \langle \chi_i, \chi_j \rangle = \delta_{ij} \quad \text{for } 1 \leq i, j \leq k. \end{align*} In particular, they are $\mathbb{C}$-linearly independent in $\mathcal{C}(G)$. Take the inner product of the equation $\chi = \sum_j m_j \chi_j$ with $\chi_\ell$ for a fixed $\ell$, using sesquilinearity of $\langle \cdot, \cdot \rangle$: \begin{align*} \langle \chi, \chi_\ell \rangle = \left\langle \sum_{j=1}^k m_j \chi_j,\; \chi_\ell \right\rangle = \sum_{j=1}^k m_j \langle \chi_j, \chi_\ell \rangle = \sum_{j=1}^k m_j \delta_{j\ell} = m_\ell. \end{align*} (The $m_j$ are real, so they are unaffected by which slot of the inner product they sit in.) Hence \begin{align*} m_\ell = \langle \chi, \chi_\ell \rangle \quad \text{for } \ell = 1, \ldots, k. \end{align*} Identically, applied to $\rho'$: \begin{align*} m'_\ell = \langle \chi', \chi_\ell \rangle \quad \text{for } \ell = 1, \ldots, k. \end{align*} [guided] The Row Orthogonality relation says the irreducible characters form an orthonormal family — they are unit length and pairwise orthogonal. We use this to "project" $\chi$ onto each $\chi_\ell$. Concretely: $\chi = \sum_j m_j \chi_j$ is a linear combination of the $\chi_j$, and we can read off the coefficient $m_\ell$ by computing $\langle \chi, \chi_\ell \rangle$. By linearity in the first slot: \begin{align*} \langle \chi, \chi_\ell \rangle = \sum_j m_j \langle \chi_j, \chi_\ell \rangle. \end{align*} Orthonormality kills every term except $j = \ell$, leaving $m_\ell \cdot 1 = m_\ell$. This is the **multiplicity formula in character form**: the multiplicity of $\rho_\ell$ in $\rho$ is precisely the inner product $\langle \chi, \chi_\ell \rangle$, computable directly from the character. (This is also a corollary of the [Multiplicity Formula](/theorems/2420) — the dimension of $\operatorname{Hom}_G(V_\ell, V)$ specialised via the character.) [/guided] [/step] [step:Conclude $m_\ell = m'_\ell$ for all $\ell$ and hence $\rho \cong \rho'$] By hypothesis, $\chi = \chi'$ as functions on $G$. Hence for every $\ell$, \begin{align*} m_\ell = \langle \chi, \chi_\ell \rangle = \langle \chi', \chi_\ell \rangle = m'_\ell. \end{align*} The multiplicities of every irreducible $\rho_\ell$ in $\rho$ and $\rho'$ coincide. From Step 2, $\rho \cong \bigoplus_j m_j \rho_j$ and $\rho' \cong \bigoplus_j m'_j \rho_j$ as $G$-representations. Since $m_j = m'_j$ for all $j$, \begin{align*} \rho \cong \bigoplus_{j=1}^k m_j \rho_j \cong \bigoplus_{j=1}^k m'_j \rho_j \cong \rho'. \end{align*} The first and third isomorphisms are by Maschke's decompositions; the middle isomorphism uses that direct sums of $G$-representations are determined up to isomorphism by the multiset of isomorphism classes of summands, with multiplicities $m_j = m'_j$. Hence $\rho \cong \rho'$ as $G$-representations. [guided] We have just shown that the character determines all the multiplicities. Now we put the pieces together. Both $\rho$ and $\rho'$ have a Maschke decomposition $\bigoplus_j m_j \rho_j$ vs $\bigoplus_j m'_j \rho_j$ (different multiplicities a priori). We have shown $m_j = m'_j$ for every $j$ — every irreducible appears with the same multiplicity in both. Two direct sums $\bigoplus_j V_j^{\oplus m_j}$ and $\bigoplus_j V_j^{\oplus m'_j}$ are isomorphic when $m_j = m'_j$ for all $j$ — pair up the $m_j$ copies of $V_j$ on the left with the $m'_j = m_j$ copies on the right by any bijection, and assemble into a $G$-equivariant isomorphism. So $\rho \cong \rho'$. The equivalence "characters equal $\iff$ representations isomorphic" thus reduces, in the strong form of Maschke, to the statement "characters of irreducibles are linearly independent". Without complete reducibility we could not even write the multiplicity decomposition; without orthogonality of irreducible characters we could not extract the multiplicities from the character. [/guided] [/step] [step:Combine both directions] The forward direction (Step 1) and the converse (Steps 2--4) together give: two complex representations $\rho, \rho'$ of $G$ are isomorphic if and only if their characters $\chi, \chi'$ are equal as functions on $G$. This is the statement of the theorem. [/step]