# Abelian-to-normal replacement theorem for prime-cube index for odd prime

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## History

The result appears to have been first proved in a paper by Alperin in 1965 and later proved as part of a larger array of results in a paper by David Jonah and Marc Konvisser in 1975.

## Statement

Suppose $p$ is an odd prime. Then, if a finite $p$-group contains an abelian subgroup of index $p^3$, it contains an abelian normal subgroup (hence, an Abelian normal subgroup of group of prime power order (?)) of index $p^3$.

## Proof

### The Jonah-Konvisser proof

Given: An odd prime $p$, a finite $p$-group $P$, an abelian subgroup $A$ of $P$ of index $p^3$.

To prove: $P$ has an abelian normal subgroup of index $p^3$.

Proof:

1. Let $M$ be a maximal subgroup of $P$ containing $A$: Such a $M$ exists and has index $p$ in $P$.
2. By fact (1), the number of abelian subgroups of $M$ of index $p^2$ is either equal to $2$ or congruent to $1$ modulo $p$.
3. If $M$ has exactly two abelian subgroups of index $p^2$, then both of them are normal in $P$: Since $M$ is maximal in $P$, it is normal (See fact (2)). Thus, any inner automorphism of $P$ sends subgroups of $M$ to subgroups of $M$. Since the sizes of orbits are either $1$ or powers of $p$, and there are only two abelian subgroups of index $p^2$, they must both be normal in $P$.
4. If the number of abelian subgroups of $M$ of index $p^2$ is congruent to $1$ modulo $p$, then there exists a subgroup of index $p^2$ in $M$ that is normal in $P$: The inner automorphisms of $P$ permute abelian subgroups of index $p^2$ in $M$, since $M$ is normal in $P$. All orbits have size either $1$ or a nontrivial power (and hence, multiple) of $p$. Since the total number is $1$ modulo $p$, there must be at least one orbit of size one.