Math-1221 · mathematics · Group theory

Math-1221

Kourovka Notebook Problem 21.51

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Statement

Definitions and mathematical statement

Fix a prime $p$ and a positive integer $n$. A finite $p$-group is a group of order $p^a$ for some integer $a\ge0$. An abelian subgroup has commuting elements. An elementary abelian $p$-subgroup is a subgroup isomorphic to $(\mathbb Z/p\mathbb Z)^r$ for some $r$, or equivalently an abelian subgroup in which every nonidentity element has order $p$. Normality of a subgroup means invariance under conjugation by every element of the ambient group.

Part (a) asks for precisely those pairs $(p,n)$ with the property \[ \forall P\text{ finite }p\text{-group},\quad [\exists A\le P\text{ abelian},\ |A|=p^n] \Longrightarrow [\exists B\trianglelefteq P\text{ abelian},\ |B|=p^n]. \] Part (b) asks for the analogous classification when both occurrences of "abelian" are replaced by "elementary abelian." For a fixed prime, this is a determination of all positive integers $n$ for which the corresponding implication is universal in $P$.

The normal subgroup supplied in the conclusion need not contain the original subgroup, be contained in it, or be conjugate to it. It must have exactly the same order, not merely some smaller nontrivial order. No bound is imposed on $|P|$ beyond finiteness and its prime-power form. A negative answer for a pair requires a $p$-group with the specified subgroup but no normal subgroup of that type and order. The two classifications are separate requirements.

Short English statement

For which primes and subgroup orders does existence of an abelian, or elementary abelian, subgroup force a normal subgroup of the same type and order?

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