Group of prime power order having a larger abelianization than any proper subgroup has class two

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Let P be a group of prime power order with the property that the order of the abelianization of P (i.e., the quotient of P by its commutator subgroup) is larger than that of any proper subgroup of P. Then, P is a Group of nilpotency class two (?).

Related facts

Facts used

  1. Third isomorphism theorem
  2. Formula for commutator of element and product of two elements


We prove this for any prime p by assuming P to be a minimal counterexample for that prime p. In other words, we show that if P is a finite p-group having a larger abelianization than every proper subgroup but does not have class two, we have a contradiction.

  1. Let N be a minimal normal subgroup of P contained in P' = [P,P]: Note that such a N exists because [P,P] is a nontrivial normal subgroup of P.
  2. Let H be a proper subgroup of P containing N. Then, the abelianization of P/N is larger in size than the abelianization of H/N: We have [(P/N):(P/N)'] = [P:P'] because N \le P' (using fact (1)). [P:P'] > [H:H'] by assumption. Finally, [H:H'] \ge [(H/N):(H/N)'] (equality need not hold since N is not necessarily contained in H'. Combining all these, [(P/N):(P/N)'] > [(H/N):(H/N)'].
  3. N = [[P,P],P]: From step (2), we know that P/N has a larger abelianization than any proper subgroup. By the minimal counterexample nature of P, P/N must have class two. Thus, [[P,P],P] \le N. But [[P,P],P] is a normal subgroup of P and N is minimal normal. Moreover, P has class greater than two, so [[P,P],P] is nontrivial. This forces N = [[P,P],P].
  4. Let C = C_P(P') and D = P'/(P' \cap Z(P)). Then, there is a map (G/C) \times D \to N given by (gC,x) \mapsto [g,x], and this map is homomorphic in each argument: Using fact (2), we see that the map is well-defined, and homomorphic in both arguments.


Journal references