[proofplan] We argue by contradiction: assume $G$ is non-abelian simple and $\mathcal{C}$ is a conjugacy class of size $p^r$ with $r \geq 1$, and pick $1 \neq g \in \mathcal{C}$. We show that for every non-trivial irreducible character $\chi$ of $G$, either $p \mid \chi(1)$ or $\chi(g) = 0$ (the [Coprimality Lemma](/theorems/2467) gives the dichotomy and faithfulness of irreducibles on a non-abelian simple group rules out the equality $|\chi(g)| = \chi(1)$). Plugging this dichotomy into the column-orthogonality identity for the columns indexed by $1$ and $g$ produces a non-integer rational number ($-1/p$) that is also an algebraic integer. The standard fact that rational algebraic integers are ordinary integers gives the contradiction. [/proofplan] [step:Set up the proof by contradiction and split irreducible characters into two cases] Assume for contradiction that $G$ is a non-abelian simple group, and let $\mathcal{C}$ be a conjugacy class of size $|\mathcal{C}| = p^r$ with $r \geq 1$. Fix $g \in \mathcal{C}$. Since $|\mathcal{C}| > 1$, $g$ does not commute with every element of $G$, so $g \notin Z(G)$. In particular $g \neq 1$ (as $1 \in Z(G)$). Let $\operatorname{Irr}(G) = \{\chi_1 = \mathbf{1}_G, \chi_2, \ldots, \chi_h\}$ be the irreducible characters of $G$. For each non-trivial $\chi \in \operatorname{Irr}(G)$, we partition into two cases according to $\gcd(\chi(1), |\mathcal{C}|) = \gcd(\chi(1), p^r)$: - **Case A:** $p \mid \chi(1)$. - **Case B:** $p \nmid \chi(1)$, equivalently $\gcd(\chi(1), p^r) = 1$. In Case B, the [Coprimality and Character Values Lemma](/theorems/2467) applies to the pair $(\chi, \mathcal{C})$. Its hypotheses — $\chi$ irreducible and $\gcd(\chi(1), |\mathcal{C}|) = 1$ — are satisfied, so \begin{align*} \chi(g) = 0 \quad \text{or} \quad |\chi(g)| = \chi(1). \end{align*} [/step] [step:Rule out $|\chi(g)| = \chi(1)$ in Case B using simplicity and non-abelianness] We show that the alternative $|\chi(g)| = \chi(1)$ in Case B cannot occur for $g \neq 1$ when $G$ is non-abelian simple. Let $\rho: G \to \operatorname{GL}(V)$ be an irreducible representation affording the non-trivial $\chi$, with $\dim V = \chi(1)$. The matrix $\rho(g) \in \operatorname{GL}(V)$ has finite order (since $g$ does), hence is diagonalisable over $\mathbb{C}$ with eigenvalues $\lambda_1, \ldots, \lambda_{\chi(1)}$ being roots of unity (in particular $|\lambda_i| = 1$). By the equality case of the triangle inequality applied to $\chi(g) = \sum_{i=1}^{\chi(1)} \lambda_i$, \begin{align*} |\chi(g)| = \chi(1) \iff \left| \sum_{i=1}^{\chi(1)} \lambda_i \right| = \sum_{i=1}^{\chi(1)} |\lambda_i| \iff \lambda_1 = \lambda_2 = \cdots = \lambda_{\chi(1)} =: \lambda, \end{align*} where the last equivalence uses that the triangle inequality is an equality for unit-modulus complex numbers exactly when they all share a common argument. Therefore $\rho(g) = \lambda I_V$ acts as a scalar. [claim:The kernel of $\rho$ is trivial] Since $\chi$ is non-trivial, $\rho$ is non-trivial, so $\ker \rho \neq G$. By definition $\ker \rho \trianglelefteq G$ is a normal subgroup. Since $G$ is simple, the only normal subgroups are $\{1\}$ and $G$. Excluding $G$, we have $\ker \rho = \{1\}$. So $\rho$ is injective. [/claim] [claim:The element $g$ lies in $Z(G)$] We have $\rho(g) = \lambda I_V$. For any $h \in G$, \begin{align*} \rho(gh) = \rho(g)\rho(h) = (\lambda I_V) \rho(h) = \rho(h)(\lambda I_V) = \rho(h)\rho(g) = \rho(hg). \end{align*} By injectivity of $\rho$ (the previous claim), $gh = hg$ for all $h \in G$, i.e.\ $g \in Z(G)$. [/claim] [claim:$Z(G) = \{1\}$ since $G$ is non-abelian simple] $Z(G) \trianglelefteq G$ is normal. By simplicity, $Z(G) \in \{\{1\}, G\}$. If $Z(G) = G$, then $G$ is abelian, contradicting non-abelianness. Hence $Z(G) = \{1\}$. [/claim] Combining the two claims: $g \in Z(G) = \{1\}$, so $g = 1$. But $g \in \mathcal{C}$ has $|\mathcal{C}| = p^r > 1$, so $g \neq 1$. Contradiction. Therefore $|\chi(g)| = \chi(1)$ does not occur, and Case B forces $\chi(g) = 0$. [guided] The structure of this step is: the [Coprimality Lemma](/theorems/2467) gives a dichotomy "$\chi(g) = 0$ or $|\chi(g)| = \chi(1)$"; we eliminate the second alternative under the structural hypothesis that $G$ is non-abelian simple. The triangle inequality equality case is the key observation. A general sum of unit complex numbers $\sum \lambda_i$ has $|\sum \lambda_i| \leq \sum |\lambda_i| = \chi(1)$, with equality iff all $\lambda_i$ point in the same direction, i.e.\ are equal. This is a generic fact, but it has a strong consequence in our context: $\rho(g)$ has all eigenvalues equal, so it is a scalar matrix. A scalar matrix $\lambda I_V$ commutes with every other matrix. So $\rho(g)$ commutes with $\rho(h)$ for every $h$. For a *faithful* representation $\rho$ (injective), the commutativity in $\operatorname{GL}(V)$ pulls back to commutativity in $G$, putting $g \in Z(G)$. Faithfulness of $\rho$ comes from simplicity of $G$ — the kernel of $\rho$ is normal, hence either trivial or all of $G$, and the latter is excluded by non-triviality of $\chi$. Finally, simplicity excludes $Z(G) = G$ unless $G$ is abelian, which is excluded by hypothesis. So $Z(G) = \{1\}$, putting $g = 1$ — contradicting $g \in \mathcal{C}$ with $|\mathcal{C}| > 1$. This is a beautiful chain: triangle equality forces scalar action, scalar action forces $g$ central (via faithfulness), centrality combined with simplicity forces $g = 1$. Each link uses one of the structural hypotheses (irreducibility, simplicity, non-abelianness) exactly once. [/guided] [/step] [step:Apply column orthogonality to the columns indexed by $1$ and $g$] By the [Column Orthogonality Relations](/theorems/2431), for $g \neq 1$, \begin{align*} 0 = \sum_{\chi \in \operatorname{Irr}(G)} \chi(1)\, \overline{\chi(g)}. \end{align*} (The relation states $\sum_\chi \chi(g_1)\overline{\chi(g_2)} = 0$ when $g_1, g_2$ are not conjugate; here $g_1 = 1$ and $g_2 = g$ are not conjugate since $\{1\}$ is its own conjugacy class and $g \neq 1$.) The trivial character $\chi_1 = \mathbf{1}_G$ contributes $\chi_1(1)\overline{\chi_1(g)} = 1 \cdot 1 = 1$. Splitting off this term and applying the dichotomy from Steps 1-2 (for non-trivial $\chi$, either $p \mid \chi(1)$ or $\chi(g) = 0$): \begin{align*} 0 = 1 + \sum_{\substack{\chi \in \operatorname{Irr}(G) \\ \chi \neq \mathbf{1}_G}} \chi(1)\, \overline{\chi(g)} = 1 + \sum_{\substack{\chi \neq \mathbf{1}_G \\ p \mid \chi(1)}} \chi(1)\, \overline{\chi(g)}, \end{align*} where the second equality uses that all summands with $\chi(g) = 0$ vanish (so the only surviving non-trivial $\chi$ are those in Case A: $p \mid \chi(1)$). [/step] [step:Derive a contradiction via the rational $\to$ integer principle] Rearranging the identity from Step 3, \begin{align*} -1 = \sum_{\substack{\chi \neq \mathbf{1}_G \\ p \mid \chi(1)}} \chi(1)\, \overline{\chi(g)}. \end{align*} For each $\chi$ in the sum, $p \mid \chi(1)$, so $\chi(1)/p \in \mathbb{Z}_{>0}$. Dividing the identity by $p$: \begin{align*} -\frac{1}{p} = \sum_{\substack{\chi \neq \mathbf{1}_G \\ p \mid \chi(1)}} \frac{\chi(1)}{p}\, \overline{\chi(g)}. \end{align*} We verify the right-hand side is an algebraic integer. For each $\chi$ in the sum: - $\chi(1)/p \in \mathbb{Z}$ is an algebraic integer (it is an ordinary integer). - $\overline{\chi(g)}$ is an algebraic integer: $\chi(g)$ is an algebraic integer by [Character Values are Algebraic Integers](/theorems/2461) (since $G$ is finite), and complex conjugation sends algebraic integers to algebraic integers (if $\beta$ is a root of $X^d + c_{d-1}X^{d-1} + \cdots + c_0 \in \mathbb{Z}[X]$, then $\bar{\beta}$ is a root of the same polynomial since $c_i \in \mathbb{Z} \subseteq \mathbb{R}$). Each summand is the product of two algebraic integers, hence an algebraic integer (algebraic integers form a subring of $\mathbb{C}$). The finite sum is therefore an algebraic integer. But the left-hand side equals $-1/p$, a rational number. By the standard fact that any rational number that is also an algebraic integer must lie in $\mathbb{Z}$ (proved as in Theorem 2464: clear denominators in a monic integer polynomial relation), we would need $-1/p \in \mathbb{Z}$. Since $p$ is a prime, $p \geq 2$, and $-1/p \notin \mathbb{Z}$. Contradiction. Therefore the original assumption — that $G$ is non-abelian simple with a class of prime-power size $p^r > 1$ — is false. So $G$ is not non-abelian simple. [guided] The contradiction comes from the fundamental incompatibility: \begin{align*} \text{rational with denominator } p \quad \text{vs.} \quad \text{algebraic integer.} \end{align*} The set $\mathbb{Q} \cap \overline{\mathbb{Z}} = \mathbb{Z}$ (rational algebraic integers are ordinary integers). So if we can simultaneously force a number to be both rational with non-trivial denominator and an algebraic integer, we have a contradiction. The column orthogonality identity manufactures such a number. The key point: the trivial character contributes $1$ to the sum, the non-trivial characters in Case B contribute $0$, and the non-trivial characters in Case A contribute terms divisible by $p$ (in the algebraic-integer sense — they have an explicit factor of $\chi(1)$, divisible by $p$). Bringing the $1$ to the other side and dividing by $p$ leaves $-1/p$ on one side and a sum of algebraic integers on the other. Discrete-vs-continuous tension: $-1/p$ has the wrong arithmetic for an algebraic integer. Why does this argument fail for abelian simple groups (e.g.\ $\mathbb{Z}/p\mathbb{Z}$)? Step 2 used non-abelianness to force $Z(G) = \{1\}$ — for abelian $G$, $Z(G) = G$, and the case $|\chi(g)| = \chi(1)$ is no longer ruled out. So the dichotomy "$\chi(g) = 0$ or $|\chi(g)| = \chi(1)$" no longer collapses to just $\chi(g) = 0$, and the column-orthogonality computation does not produce $-1/p$ on the LHS. Why does Step 2 require *both* simplicity and non-abelianness? Simplicity makes $\rho$ faithful and rules out $Z(G) \neq \{1, G\}$; non-abelianness then closes the loophole $Z(G) = G$. Together, they force $Z(G) = \{1\}$ and the contradiction $g \in Z(G) \setminus \{1\}$. This argument is the entry point to Burnside's $p^a q^b$ theorem (the only finite simple groups with two prime divisors are cyclic of prime order), one of the gems of character theory. [/guided] [/step]