The Brauer correspondent of a block

Fix an actual defect group $D$ of a block $B$ of $kG$. By Brauer's First Main Theorem, there is a unique block $b$ of $k{N}_G(D)$ having defect group $D$ and satisfying $b^G=B$. It is the Brauer correspondent of $B$ at $D$. Existence and uniqueness are the already proved bijection, so the notation does not choose an arbitrary block.

Replacing $D$ by $gD{g}^{-1}$ transports $b$ to ${}^{g}b$, by the theorem's equivariance. Thus the actual subgroup is part of the input, and passing to a defect conjugacy class gives the construction only up to the corresponding conjugation. If $N_G(D)=G$, including $D=1$, the correspondent is $B$ itself. The block bijection used here has a choice-free proof; no AC is needed for this definition.