Question

Assume char k = 2. Let A be an abelian 2'-group and let G be the semidirect product of A and a cyclic group <x> of order 2, where x acts on A by a → a^−1. If A is infinite, show that kG has no nonzero nil ideals.

Answer 1

First note that, $kA$ has no nonzero nil ideals. Since $kA$ is commutative, this simply means that the only nilpotent element of $kA$ is zero. For $\alpha = \sum_{a \in A} \alpha_a a \in kA$, let $\alpha^* = \sum \alpha_a a^{-1}$. This defines an involution on $kA$, with $x\alpha = \alpha^* x$ for any $\alpha \in kA$. Any element $\sigma \in kG$ can be expressed uniquely in the form $\alpha + \beta x$, with $\alpha, \beta \in kA$. Let $I$ be any nil ideal in $kG$, and let $\sigma \in I$. Then $(\alpha + \beta x)(\alpha^* + x\beta^*) = \alpha\alpha^* + \beta\beta^* + (\beta\alpha + \alpha\beta)x = \alpha\alpha^* + \beta\beta^* \in kA$. Since this element is nilpotent, we must have $\alpha\alpha^* = \beta\beta^*$.

Therefore,

$(\alpha + \beta x)^2 = \alpha^2 + \beta x \beta x + \alpha \beta x + \beta x \alpha = \alpha^2 + \beta \beta^* + (\alpha \beta + \beta \alpha^*) x = \alpha^2 + \alpha \alpha^* + (\alpha \beta + \alpha^* \beta) x = (\alpha + \alpha^*) (\alpha + \beta x).$

Say $(\alpha + \beta x)^{\alpha} = 0$.

Then we have $0 = (\alpha + \alpha^*)^{n-1}(\alpha + \beta x)$, so that $(\alpha + \alpha^*)^{n-1}\alpha = 0$. But then $0 = x[(\alpha + \alpha^*)^{n-1}\alpha]x = (\alpha + \alpha^*)^{n-1}\alpha^*$.

Therefore, by addition, $(\alpha + \alpha^*)^\alpha = 0$, so $\alpha = \alpha^*$. Now consider any $b \in A$. Then $b(\alpha + \beta x) \in I$ implies $b\alpha = (b\alpha)^*$. Suppose $\alpha \neq 0$; say $\alpha$ involves some group element $b - 1 \in A$. Then $1 \in \operatorname{supp}(b\alpha)$, and $b\alpha = (b\alpha)^*$ implies that $|\operatorname{supp}(\alpha)| = |\operatorname{supp}(b\alpha)|$ is odd, since $A$ has no element of order 2. But if $\operatorname{supp}(\alpha)$ misses some element $c - 1 \in A$, then $1 \notin \operatorname{supp}(c\alpha)$, and $c\alpha = (c\alpha)^*$ would imply that $|\operatorname{supp}(\alpha)| = |\operatorname{supp}(c\alpha)|$ is even. Therefore, we must have $\operatorname{supp}(\alpha) = A$. If $A$ is infinite, this is impossible. In this case, we conclude that $\alpha = 0$, and since $\sigma x = \beta x^2 = \beta$ is nilpotent, $\beta = 0$ too, so $\sigma = 0$. This completes the proof.