# Powering map by field characteristic is same in algebra and algebra group

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## Statement

Suppose $F$ is a field of characteristic equal to a prime number $p$ and $N$ is an associative algebra over $F$ that admits an algebra group $G = \{ 1 + x \mid x \in N \}$. (This happens, for instance, if $N$ is nilpotent). Then, the $p$-power map in $N$ corresponds to the $p$-power map in $G$, i.e.:

$(1 + x)^p = 1 + x^p \ \forall \ x \in N$

## Proof

The proof simply involves expanding using the binomial theorem and then using characteristic $p$. Note that we crucially use the fact that 1 commutes with $x$ to apply the binomial theorem.