Nilpotent group: Difference between revisions

; (b)→(c): Let ''p''₁,''p''₂,...,''p''_''s'' be the distinct primes dividing its order and let ''P''_''i'' in ''Syl''_{''p''_''i''}(''G''), 1 ≤ ''i'' ≤ ''s''. Let ''P'' = ''P''_''i'' for some ''i'' and let ''N'' = ''N''_''G''(''P''). Since ''P'' is a normal Sylow subgroup of ''N'', ''P'' is [[characteristic subgroup|characteristic]] in ''N''. Since ''P'' char ''N'' and ''N'' is a normal subgroup of ''N''_''G''(''N''), we get that ''P'' is a normal subgroup of ''N''_''G''(''N''). This means ''N''_''G''(''N'') is a subgroup of ''N'' and hence ''N''_''G''(''N'') = ''N''. By (b) we must therefore have ''N'' = ''G'', which gives (c).

; (c)→(d): Let ''p''₁,''p''₂,...,''p''_''s'' be the distinct primes dividing its order and let ''P''_''i'' in ''Syl''_{''p''_''i''}(''G''), 1 ≤ ''i'' ≤ ''s''. For any ''t'', 1 ≤ ''t'' ≤ ''s'' we show inductively that ''P''₁''P''₂···''P''_''t'' is isomorphic to ''P''₁×''P''₂×···×''P''_''t''. {{paragraph}}Note first that each ''P''_''i'' is normal in ''G'' so ''P''₁''P''₂···''P''_''t'' is a subgroup of ''G''. Let ''H'' be the product ''P''₁''P''₂···''P''_''t''−1 and let ''K'' = ''P''_''t'', so by induction ''H'' is isomorphic to ''P''₁×''P''₂×···×''P''_''t''−1. In particular,|''H''| = |''P''₁|⋅|''P''₂|⋅···⋅|''P''_''t''−1|. Since |''K''| = |''P''_''t''|, the orders of ''H'' and ''K'' are relatively prime. Lagrange's Theorem implies the intersection of ''H'' and ''K'' is equal to 1. By definition,''P''₁''P''₂···''P''_''t'' = ''HK'', hence ''HK'' is isomorphic to ''H''×''K'' which is equal to ''P''₁×''P''₂×···×''P''_''t''. This completes the induction. Now take ''t'' = ''s'' to obtain (d).

; (e)→(a): For any prime ''p'' dividing |''G''|, the [[Sylow group|Sylow ''p''-subgroup]] is normal. Thus we can apply (c) (since we already proved (c)→(e)).