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

## Statement

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 (?).

## Proof

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.