If $|G|=2673=3^511$ then let $S \in Syl_3(G)$. Then the number of conjugates of $S$, $n_s \equiv 1$ mod $3$ and $n_s | 11$. But the only divisors of 11 are 1 and 11, and therefore $n_s=1$ so the $S$ is the unique sylow-3 subgroup of G and hence is normal. Is this correct?
2026-03-25 22:04:25.1774476265
Proving every group of order 2673 has a non-trivial proper normal subgroup
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