Group of prime power order having a larger abelianization than any proper subgroup has class two
From Groupprops
Statement
Let be a group of prime power order with the property that the order of the abelianization of (i.e., the quotient of by its commutator subgroup) is larger than that of any proper subgroup of . Then, is a Group of nilpotency class two (?).
Related facts
- Thompson's replacement theorem for abelian subgroups
- Thompson's replacement theorem for elementary abelian subgroups
Facts used
Proof
We prove this for any prime by assuming to be a minimal counterexample for that prime . In other words, we show that if is a finite -group having a larger abelianization than every proper subgroup but does not have class two, we have a contradiction.
- Let be a minimal normal subgroup of contained in : Note that such a exists because is a nontrivial normal subgroup of .
- Let be a proper subgroup of containing . Then, the abelianization of is larger in size than the abelianization of : We have because (using fact (1)). by assumption. Finally, (equality need not hold since is not necessarily contained in . Combining all these, .
- : From step (2), we know that has a larger abelianization than any proper subgroup. By the minimal counterexample nature of , must have class two. Thus, . But is a normal subgroup of and is minimal normal. Moreover, has class greater than two, so is nontrivial. This forces .
- Let and . Then, there is a map given by , and this map is homomorphic in each argument: Using fact (2), we see that the map is well-defined, and homomorphic in both arguments.
References
Journal references
- A replacement theorem for p-groups and a conjecture by John Griggs Thompson, Journal of Algebra, ISSN 00218693, Volume 13, Page 149 - 151(Year 1969): In this paper, Thompson proved the replacement theorem for abelian subgroups, as well as the result that group of prime power order having a larger abelianization than any proper subgroup has class two.^{Official copy}^{More info}