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in the proof of the first lemma, why is $hns=\phi(hs)$ true? I simply cannot get it...
2026-04-04 13:39:36.1775309976
Question about May's Algebraic Topology book
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$\phi$ is an automorphism of the $G$-set $S$, i.e. $\phi: S \to S$ is a bijection with $\phi(gt) = g\phi(t)$ for all $g \in G$, $t \in S$. For our particular choice of $s$ in the lemma we know $\phi(s) = ns$, so $\phi(hs) = h\phi(s) = hns$ (since $h \in H \subset G$).