[proofplan] We proceed by strong induction on the number of characters $m$. The base case $m = 1$ is immediate: a single nonzero character takes values in $L^\times$, so the only scalar annihilating it everywhere is $\lambda_1 = 0$. For the inductive step with $m \ge 2$ characters, we assume a nontrivial relation $\sum_{i=1}^{m} \lambda_i \phi_i(g) = 0$ for all $g \in G$ and derive a contradiction. The key manoeuvre exploits the distinctness of $\phi_1$ and $\phi_m$: substituting $gh$ for $g$ (where $h \in G$ is chosen so that $\phi_1(h) \neq \phi_m(h)$) produces a second relation, and subtracting an appropriate scalar multiple of the original from it eliminates the $\phi_m$ term, yielding a nontrivial relation among fewer than $m$ characters — contradicting the inductive hypothesis. [/proofplan] [step:Handle the base case $m = 1$] Suppose $m = 1$ and $\lambda_1 \in L$ satisfies $\lambda_1 \phi_1(g) = 0$ for all $g \in G$. Since $\phi_1$ is a group homomorphism $\phi_1 \colon G \to L^\times$, the image $\phi_1(g)$ lies in $L^\times$ for every $g \in G$. In particular, $\phi_1(g) \neq 0$ for every $g \in G$. Multiplying both sides of $\lambda_1 \phi_1(g) = 0$ by $\phi_1(g)^{-1} \in L$ gives $\lambda_1 = 0$. [guided] The base case reduces to a simple cancellation in the field $L$. The relation $\lambda_1 \phi_1(g) = 0$ holds for all $g \in G$, and we need to conclude $\lambda_1 = 0$. Why can we cancel $\phi_1(g)$? Because $\phi_1$ maps into $L^\times$, the group of units of $L$. Every value $\phi_1(g)$ is a nonzero element of $L$, hence invertible. Choosing any particular $g \in G$ (for instance $g = e$, the identity element, which gives $\phi_1(e) = 1$), the equation becomes $\lambda_1 \cdot 1 = 0$, so $\lambda_1 = 0$. This is the only place in the proof where we directly use the fact that characters take values in $L^\times$ rather than merely in $L$. For $m \ge 2$, the argument will instead exploit the multiplicativity of characters to reduce to smaller collections. [/guided] [/step] [step:Assume a nontrivial relation among $m \ge 2$ characters and set up the inductive hypothesis] Fix $m \ge 2$ and assume, as the inductive hypothesis, that the theorem holds for any collection of fewer than $m$ distinct group homomorphisms $G \to L^\times$. That is: if $\psi_1, \ldots, \psi_r \colon G \to L^\times$ are distinct characters with $r < m$ and $\mu_1 \psi_1(g) + \cdots + \mu_r \psi_r(g) = 0$ for all $g \in G$, then $\mu_1 = \cdots = \mu_r = 0$. Suppose for contradiction that the theorem fails for $m$. Then there exist $\lambda_1, \ldots, \lambda_m \in L$, not all zero, such that \begin{align*} \lambda_1 \phi_1(g) + \lambda_2 \phi_2(g) + \cdots + \lambda_m \phi_m(g) = 0 \quad \text{for all } g \in G. \tag{$\star$} \end{align*} By relabelling if necessary, we may assume $\lambda_m \neq 0$. (At least one $\lambda_i$ is nonzero by assumption; we relabel to place it in position $m$.) [guided] This step frames the induction precisely. We use strong induction: the inductive hypothesis asserts the result for all collections of fewer than $m$ distinct characters, not merely for $m - 1$ characters. (In this proof, we will only need the case of $m - 1$ characters, but the strong induction formulation is standard.) The goal of the remaining steps is to derive a contradiction from the existence of a nontrivial relation $(\star)$ among $m$ characters. The strategy is to manufacture from $(\star)$ a new relation involving strictly fewer characters, at least one of whose coefficients is nonzero — contradicting the inductive hypothesis. Why do we ensure $\lambda_m \neq 0$? Because the elimination trick in the next step will cancel the $\phi_m$ term. If $\lambda_m$ were zero, the relation $(\star)$ would already involve fewer than $m$ characters (since the $\phi_m$ term would be absent), and the inductive hypothesis would immediately yield $\lambda_1 = \cdots = \lambda_{m-1} = 0$, contradicting our assumption that the $\lambda_i$ are not all zero. So $\lambda_m \neq 0$ is forced. [/guided] [/step] [step:Choose $h \in G$ that separates $\phi_1$ from $\phi_m$] Since $\phi_1$ and $\phi_m$ are distinct group homomorphisms $G \to L^\times$, there exists $h \in G$ such that $\phi_1(h) \neq \phi_m(h)$. [guided] This is the step where the hypothesis that the characters are **distinct** is consumed. Two functions $\phi_1, \phi_m \colon G \to L^\times$ are distinct if and only if they differ at some point of $G$, so there exists $h \in G$ with $\phi_1(h) \neq \phi_m(h)$. Why do we specifically separate $\phi_1$ from $\phi_m$, rather than some other pair? The choice is a matter of convenience: in the next step, we will form a linear combination that eliminates the $\phi_m$ term, and the coefficient of $\phi_1$ in the resulting relation will involve the factor $(\phi_1(h) - \phi_m(h))$. By choosing the pair $(\phi_1, \phi_m)$, we ensure this factor is nonzero, which guarantees the new relation is nontrivial. Any pair of distinct characters $(\phi_i, \phi_m)$ with $i \neq m$ would work equally well, but fixing $i = 1$ simplifies the bookkeeping. [/guided] [/step] [step:Substitute $gh$ for $g$ in $(\star)$ and use multiplicativity to produce a second relation] Since $(\star)$ holds for all $g \in G$, it holds in particular with $g$ replaced by $gh$ (as $h \in G$ is fixed and $gh \in G$ for every $g \in G$): \begin{align*} \lambda_1 \phi_1(gh) + \lambda_2 \phi_2(gh) + \cdots + \lambda_m \phi_m(gh) = 0 \quad \text{for all } g \in G. \end{align*} Since each $\phi_i$ is a group homomorphism, $\phi_i(gh) = \phi_i(g) \phi_i(h)$ for every $g \in G$. Substituting: \begin{align*} \lambda_1 \phi_1(h) \phi_1(g) + \lambda_2 \phi_2(h) \phi_2(g) + \cdots + \lambda_m \phi_m(h) \phi_m(g) = 0 \quad \text{for all } g \in G. \tag{$\star\star$} \end{align*} [guided] The key idea is that a functional identity that holds for all $g \in G$ also holds when we replace $g$ by $gh$, since $g \mapsto gh$ is a bijection $G \to G$ (right multiplication by a fixed element). The group homomorphism property then factors $\phi_i(gh) = \phi_i(g)\phi_i(h)$, pulling the constant $\phi_i(h)$ out of each term. We now have two identities — $(\star)$ and $(\star\star)$ — both valid for all $g \in G$. The next step combines them to cancel the $\phi_m$ term. [/guided] [/step] [step:Eliminate $\phi_m$ by subtracting $\phi_m(h)$ times $(\star)$ from $(\star\star)$] Multiply relation $(\star)$ by the scalar $\phi_m(h) \in L^\times$: \begin{align*} \lambda_1 \phi_m(h) \phi_1(g) + \lambda_2 \phi_m(h) \phi_2(g) + \cdots + \lambda_m \phi_m(h) \phi_m(g) = 0 \quad \text{for all } g \in G. \tag{$\star'$} \end{align*} Subtract $(\star')$ from $(\star\star)$. The $\phi_m$ terms cancel: the coefficient of $\phi_m(g)$ in $(\star\star)$ is $\lambda_m \phi_m(h)$ and in $(\star')$ is $\lambda_m \phi_m(h)$, so their difference is zero. For $i \in \{1, \ldots, m-1\}$, the coefficient of $\phi_i(g)$ becomes $\lambda_i \phi_i(h) - \lambda_i \phi_m(h) = \lambda_i(\phi_i(h) - \phi_m(h))$. This yields \begin{align*} \sum_{i=1}^{m-1} \lambda_i \bigl(\phi_i(h) - \phi_m(h)\bigr) \phi_i(g) = 0 \quad \text{for all } g \in G. \tag{$\dagger$} \end{align*} [guided] The subtraction trick is designed to kill the last term. In $(\star\star)$, the coefficient of $\phi_m(g)$ is $\lambda_m \phi_m(h)$. In $(\star)$ scaled by $\phi_m(h)$, the coefficient of $\phi_m(g)$ is also $\lambda_m \phi_m(h)$. Subtracting forces exact cancellation in the $m$-th position. For the remaining positions $i = 1, \ldots, m - 1$, the coefficient of $\phi_i(g)$ after subtraction is \begin{align*} \lambda_i \phi_i(h) - \lambda_i \phi_m(h) = \lambda_i \bigl(\phi_i(h) - \phi_m(h)\bigr). \end{align*} Why multiply by $\phi_m(h)$ rather than, say, dividing $(\star\star)$ by $\phi_m(h)$? Both approaches work (since $\phi_m(h) \in L^\times$ is invertible), but multiplying $(\star)$ by $\phi_m(h)$ is notationally cleaner and avoids introducing fractions. The result is a relation $(\dagger)$ involving only the $m - 1$ characters $\phi_1, \ldots, \phi_{m-1}$. The next step shows it is nontrivial, yielding the desired contradiction. [/guided] [/step] [step:Verify that the new relation $(\dagger)$ is nontrivial and apply the inductive hypothesis to reach a contradiction] Relation $(\dagger)$ is a linear dependence relation among the $m - 1$ distinct characters $\phi_1, \ldots, \phi_{m-1}$, with coefficients $\mu_i := \lambda_i(\phi_i(h) - \phi_m(h))$ for $i = 1, \ldots, m - 1$. We check that not all $\mu_i$ are zero. Consider the coefficient $\mu_1 = \lambda_1(\phi_1(h) - \phi_m(h))$. We have two cases: - If $\lambda_1 \neq 0$, then $\mu_1 \neq 0$ because $\lambda_1 \neq 0$ and $\phi_1(h) - \phi_m(h) \neq 0$ (by the choice of $h$ in the previous step), and $L$ is a field (hence an integral domain), so the product of two nonzero elements is nonzero. - If $\lambda_1 = 0$, then the original relation $(\star)$ reads $\lambda_2 \phi_2(g) + \cdots + \lambda_m \phi_m(g) = 0$ for all $g \in G$, which is a nontrivial relation (since $\lambda_m \neq 0$) among the $m - 1$ distinct characters $\phi_2, \ldots, \phi_m$, directly contradicting the inductive hypothesis. In either case, we obtain a contradiction with the inductive hypothesis: in the first case, the relation $(\dagger)$ is a nontrivial relation among $m - 1$ distinct characters $\phi_1, \ldots, \phi_{m-1}$; in the second case, the original relation $(\star)$ (with $\lambda_1 = 0$ dropped) is a nontrivial relation among $m - 1$ distinct characters $\phi_2, \ldots, \phi_m$. Both contradict the assumption that the theorem holds for fewer than $m$ characters. Therefore, no nontrivial relation among $m$ distinct characters can exist, and we conclude $\lambda_1 = \lambda_2 = \cdots = \lambda_m = 0$. [guided] This is the crux of the inductive step. We need to verify that the relation $(\dagger)$ produced in the previous step is **nontrivial** — that is, that at least one of the coefficients $\mu_i = \lambda_i(\phi_i(h) - \phi_m(h))$ is nonzero. Without nontriviality, the inductive hypothesis would give no information. Consider $\mu_1 = \lambda_1(\phi_1(h) - \phi_m(h))$. The factor $\phi_1(h) - \phi_m(h)$ is nonzero by our choice of $h$. So $\mu_1 = 0$ if and only if $\lambda_1 = 0$ (using the fact that $L$ has no zero divisors, being a field). **Case 1: $\lambda_1 \neq 0$.** Then $\mu_1 = \lambda_1(\phi_1(h) - \phi_m(h)) \neq 0$, so the relation $(\dagger)$ is a nontrivial relation among the $m - 1$ distinct characters $\phi_1, \ldots, \phi_{m-1}$. The inductive hypothesis (applied to these $m - 1$ characters) asserts that the only such relation is the trivial one — contradiction. **Case 2: $\lambda_1 = 0$.** Then we do not even need relation $(\dagger)$. The original relation $(\star)$ becomes \begin{align*} \lambda_2 \phi_2(g) + \lambda_3 \phi_3(g) + \cdots + \lambda_m \phi_m(g) = 0 \quad \text{for all } g \in G, \end{align*} with $\lambda_m \neq 0$. This is a nontrivial relation among the $m - 1$ distinct characters $\phi_2, \ldots, \phi_m$, contradicting the inductive hypothesis directly. In both cases, we reach a contradiction with the assumption that a nontrivial relation exists among $m$ characters. This completes the inductive step. It is worth noting where each hypothesis was used: - **$\phi_i$ are group homomorphisms:** in the substitution $\phi_i(gh) = \phi_i(g)\phi_i(h)$ (Step 4). - **$\phi_i$ map into $L^\times$:** in the base case, to cancel $\phi_1(g)$ (Step 1), and implicitly throughout to ensure $\phi_i(h) \in L^\times$. - **The $\phi_i$ are distinct:** in the choice of $h$ with $\phi_1(h) \neq \phi_m(h)$ (Step 3). - **$L$ is a field:** in the conclusion that $\lambda_1(\phi_1(h) - \phi_m(h)) \neq 0$ when both factors are nonzero (this step), using the absence of zero divisors. [/guided] [/step]