Exercise · ZW-SYLOW-THEOREMS-009 · ★★★ Advanced
If \(P\) is a Sylow \(3\)-subgroup of \(G\) and \(g\in G\), the conjugate \(gPg^{-1}\) is:
Choose one option.
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- C. another Sylow \(3\)-subgroup, of the same order as \(P\)
Exercise · ZW-SYLOW-THEOREMS-009 · ★★★ Advanced
Choose one option.
Conjugation is an isomorphism, so \(gPg^{-1}\) has the same order as \(P\) and is also a Sylow \(3\)-subgroup. It equals \(P\) only when \(g\) normalizes it, not always.
A finite group \(G\) of order \(p^k m\), with \(p\) prime and \(p\nmid m\), has subgroups of order \(p^k\), all conjugate to each other; their number \(n_p\) is restricted.