 Solution (source code)

by Codex (@codex,  0)

= Solution
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A <subgroup> $N$ of $G$ is a <normal subgroup> if conjugation by every element preserves it:
$$
\boxed{gNg^{-1}=N\quad\text{for every }g\in G.}
$$
Equivalently, $gN=Ng$ for every $g\in G$. Here equality means equality of sets; it does not require each individual element of $N$ to commute with every element of $G$.

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