[proofplan] The starting point is the $G$-equivariant decomposition $V \otimes V = S^2 V \oplus \Lambda^2 V$ of the tensor square into its symmetric and antisymmetric parts (valid in characteristic zero). Taking characters gives $\chi^2 = \chi_S + \chi_\Lambda$ since the character of a tensor product is a product of characters by [Tensor Product of Representations](/theorems/2441). The substantive content is the explicit formulas for $\chi_S(g)$ and $\chi_\Lambda(g)$. We diagonalise $\rho(g)$ on $V$ with eigenvalues $\lambda_1, \ldots, \lambda_n$ and read off eigenbases of $S^2 V$ and $\Lambda^2 V$ in terms of symmetrised and antisymmetrised pure tensors. The eigenvalue sums simplify via the algebraic identity $(\sum_i \lambda_i)^2 = \sum_i \lambda_i^2 + 2\sum_{i<j} \lambda_i \lambda_j$, with the squares $\sum_i \lambda_i^2 = \chi(g^2)$ identified using $\rho(g)^2 = \rho(g^2)$. Solving the linear system yields the stated formulas. [/proofplan] [step:Decompose $V \otimes V$ as $S^2 V \oplus \Lambda^2 V$ and read off $\chi^2 = \chi_S + \chi_\Lambda$] The transposition $\tau \in \operatorname{End}(V \otimes V)$ defined on pure tensors by $\tau(v \otimes w) := w \otimes v$ (extended linearly) satisfies $\tau^2 = I$, so its eigenvalues are $\pm 1$. The symmetric square $S^2 V := \ker(\tau - I)$ and the exterior square $\Lambda^2 V := \ker(\tau + I)$ are the $\pm 1$ eigenspaces: \begin{align*} V \otimes V = S^2 V \oplus \Lambda^2 V, \end{align*} with explicit projections \begin{align*} P_S = \tfrac{1}{2}(I + \tau), \qquad P_\Lambda = \tfrac{1}{2}(I - \tau) \end{align*} satisfying $P_S + P_\Lambda = I$, $P_S P_\Lambda = 0$, $P_S^2 = P_S$, $P_\Lambda^2 = P_\Lambda$ (the factor $\tfrac{1}{2}$ is valid because we work over $\mathbb{C}$, where $2$ is invertible). The decomposition is $G$-invariant. For any $g \in G$ and pure tensor, \begin{align*} \tau \circ (\rho \otimes \rho)(g)(v \otimes w) = \tau\bigl(\rho(g)v \otimes \rho(g)w\bigr) = \rho(g)w \otimes \rho(g)v = (\rho \otimes \rho)(g)(w \otimes v) = (\rho \otimes \rho)(g) \circ \tau(v \otimes w), \end{align*} so $\tau$ commutes with $(\rho \otimes \rho)(g)$. By linearity and density of pure tensors, $\tau$ commutes with $(\rho \otimes \rho)(g)$ on all of $V \otimes V$. Hence the eigenspaces $S^2 V$ and $\Lambda^2 V$ are $G$-invariant subrepresentations, and $V \otimes V = S^2 V \oplus \Lambda^2 V$ is a decomposition of representations. Taking characters and applying [Hom Additivity / Direct Sum Additivity of Characters](/theorems/2419), \begin{align*} \chi_{V \otimes V} = \chi_{S^2 V} + \chi_{\Lambda^2 V} = \chi_S + \chi_\Lambda. \end{align*} By [Tensor Product of Representations](/theorems/2441), $\chi_{V \otimes V}(g) = \chi(g)^2$. Hence \begin{align*} \chi(g)^2 = \chi_S(g) + \chi_\Lambda(g) \end{align*} for all $g \in G$, as claimed. [guided] The decomposition $V \otimes V = S^2 V \oplus \Lambda^2 V$ uses the involution $\tau: V \otimes V \to V \otimes V$ that swaps the two factors. Since $\tau^2 = I$ and we are over $\mathbb{C}$ (where $2$ is invertible), $\tau$ is diagonalisable with eigenvalues $\pm 1$ and the projections onto the eigenspaces are $\tfrac{1}{2}(I \pm \tau)$. The crucial point is that $\tau$ commutes with the diagonal $G$-action $(\rho \otimes \rho)(g) = \rho(g) \otimes \rho(g)$. This is because the action treats both factors **the same way**: applying $\rho(g)$ to $v$ and to $w$ separately, then swapping, is the same as swapping first and then applying $\rho(g)$ to each factor. Since $\tau$ commutes with the $G$-action, the eigenspaces of $\tau$ are $G$-invariant, hence subrepresentations. The character relation follows from two facts: (i) characters are additive on direct sums of representations ([Hom Additivity](/theorems/2419)), and (ii) the character of a tensor product is the product of characters ([Tensor Product of Representations](/theorems/2441)). So \begin{align*} \chi(g)^2 = \chi_{V \otimes V}(g) = \chi_{S^2 V \oplus \Lambda^2 V}(g) = \chi_S(g) + \chi_\Lambda(g). \end{align*} This identity alone gives one equation in two unknowns; we need one more equation to solve for $\chi_S$ and $\chi_\Lambda$ individually. The next step provides it by computing $\chi_\Lambda$ directly via diagonalisation. [/guided] [/step] [step:Diagonalise $\rho(g)$ and compute $\chi_\Lambda(g) = \sum_{i<j} \lambda_i \lambda_j$ on an eigenbasis of $\Lambda^2 V$] Fix $g \in G$. Since $G$ is finite, $\rho(g)$ has finite order and is diagonalisable over $\mathbb{C}$. Choose a basis $v_1, \ldots, v_n$ of $V$ with $\rho(g)v_i = \lambda_i v_i$, where $\lambda_1, \ldots, \lambda_n$ are roots of unity. The induced basis on $V \otimes V$ is $\{v_i \otimes v_j : 1 \leq i, j \leq n\}$, and from Step 1's diagonalisation \begin{align*} (\rho \otimes \rho)(g)(v_i \otimes v_j) = \lambda_i \lambda_j\, (v_i \otimes v_j). \end{align*} For the exterior square, write $v_i \wedge v_j := \tfrac{1}{2}(v_i \otimes v_j - v_j \otimes v_i)$ — the antisymmetric projection of $v_i \otimes v_j$. The set $\{v_i \wedge v_j : 1 \leq i < j \leq n\}$ forms a basis of $\Lambda^2 V$, of dimension $\binom{n}{2}$. Each is an eigenvector of $(\rho \otimes \rho)(g)$: \begin{align*} (\rho \otimes \rho)(g)(v_i \wedge v_j) &= \tfrac{1}{2}\bigl(\lambda_i \lambda_j\, v_i \otimes v_j - \lambda_j \lambda_i\, v_j \otimes v_i\bigr) \\ &= \lambda_i \lambda_j \cdot \tfrac{1}{2}(v_i \otimes v_j - v_j \otimes v_i) \\ &= \lambda_i \lambda_j\, (v_i \wedge v_j). \end{align*} So the action of $g$ on $\Lambda^2 V$ is diagonal in the basis $\{v_i \wedge v_j : i < j\}$ with eigenvalues $\lambda_i \lambda_j$. Summing the eigenvalues: \begin{align*} \chi_\Lambda(g) = \sum_{1 \leq i < j \leq n} \lambda_i \lambda_j. \end{align*} [/step] [step:Apply the algebraic identity $(\sum_i \lambda_i)^2 = \sum_i \lambda_i^2 + 2 \sum_{i<j} \lambda_i \lambda_j$ to extract $\chi_\Lambda$] Squaring $\chi(g) = \sum_{i=1}^n \lambda_i$: \begin{align*} \chi(g)^2 = \biggl(\sum_{i=1}^n \lambda_i\biggr)^2 = \sum_{i=1}^n \lambda_i^2 + 2 \sum_{1 \leq i < j \leq n} \lambda_i \lambda_j. \end{align*} The first sum on the right is identified using $\rho$ a homomorphism: $\rho(g)^2 = \rho(g \cdot g) = \rho(g^2)$. Since $\rho(g)$ is diagonal with eigenvalues $\lambda_i$ in the basis $\{v_i\}$, $\rho(g)^2$ is diagonal in the same basis with eigenvalues $\lambda_i^2$. Therefore \begin{align*} \chi(g^2) = \operatorname{tr}(\rho(g^2)) = \operatorname{tr}(\rho(g)^2) = \sum_{i=1}^n \lambda_i^2. \end{align*} The second sum on the right is $2 \chi_\Lambda(g)$ by Step 2. Substituting: \begin{align*} \chi(g)^2 = \chi(g^2) + 2 \chi_\Lambda(g), \end{align*} and solving for $\chi_\Lambda(g)$: \begin{align*} \chi_\Lambda(g) = \tfrac{1}{2}\bigl(\chi(g)^2 - \chi(g^2)\bigr). \end{align*} Combining with $\chi(g)^2 = \chi_S(g) + \chi_\Lambda(g)$ from Step 1: \begin{align*} \chi_S(g) = \chi(g)^2 - \chi_\Lambda(g) = \chi(g)^2 - \tfrac{1}{2}\bigl(\chi(g)^2 - \chi(g^2)\bigr) = \tfrac{1}{2}\bigl(\chi(g)^2 + \chi(g^2)\bigr). \end{align*} Both formulas are established for arbitrary $g \in G$. [guided] The two formulas \begin{align*} \chi_S(g) = \tfrac{1}{2}\bigl(\chi(g)^2 + \chi(g^2)\bigr), \qquad \chi_\Lambda(g) = \tfrac{1}{2}\bigl(\chi(g)^2 - \chi(g^2)\bigr) \end{align*} are extracted from a $2 \times 2$ linear system in the unknowns $\chi_S(g)$ and $\chi_\Lambda(g)$: \begin{align*} \chi_S(g) + \chi_\Lambda(g) &= \chi(g)^2, \\ \chi_S(g) - \chi_\Lambda(g) &= \chi(g^2). \end{align*} The first equation is from Step 1 — the character of $V \otimes V$ split as $S^2 V + \Lambda^2 V$. The second equation comes from the eigenvalue identity proved in Steps 2 and 3. To derive the second equation, the key algebraic identity is \begin{align*} \biggl(\sum_i \lambda_i\biggr)^2 = \sum_i \lambda_i^2 + 2 \sum_{i<j} \lambda_i \lambda_j. \end{align*} This is the standard expansion of a squared sum into diagonal and off-diagonal parts. The off-diagonal part is $2\chi_\Lambda(g)$ by the diagonalisation in Step 2 (each unordered pair $\{i,j\}$ appears once in $\sum_{i<j}$). The diagonal part is $\chi(g^2)$ by the homomorphism property: $\rho(g^2) = \rho(g)^2$, which on the eigenbasis acts diagonally with eigenvalues $\lambda_i^2$. Symbolically, $\chi(g)^2$ packages $S^2 + \Lambda^2$, and $\chi(g^2)$ packages $S^2 - \Lambda^2$ (one can verify this by an analogous diagonalisation on $S^2 V$, using basis $v_i v_j$ for $i \leq j$ with eigenvalues $\lambda_i \lambda_j$, giving $\chi_S(g) = \sum_{i \leq j} \lambda_i \lambda_j$, and observing $\chi_S - \chi_\Lambda = \sum_i \lambda_i^2 = \chi(g^2)$). Adding and subtracting recovers the formulas. The appearance of $\chi(g^2)$ is the substantive structural feature: the character of a symmetric or exterior power involves not only the original character but also values at iterated powers. This pattern generalises: $\chi_{S^k V}(g)$ and $\chi_{\Lambda^k V}(g)$ are polynomial in $\chi(g), \chi(g^2), \ldots, \chi(g^k)$ via Newton's identities applied to power sums of eigenvalues. [/guided] [/step]