Baer norm not is hereditarily normal

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., Baer norm) does not always satisfy a particular subgroup property (i.e., hereditarily normal subgroup)
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Statement

The Baer norm of a group need not be a hereditarily normal subgroup: not every subgroup of the Baer norm need be normal in the whole group.

Definitions used

Baer norm

Further information: Baer norm

The Baer norm of a group is defined as the intersection of the normalizers of all its subgroups.

Related facts

• Baer norm is hereditarily permutable
• Baer norm is Dedekind: Every subgroup of the Baer norm is normal within the Baer norm.
• Normality is not transitive: Combining the fact that the Baer norm is normal, every subgroup of the Baer norm is normal in it, and not every subgroup of the Baer norm is normal in the whole group, gives a proof that normal subgroups of normal subgroups need not be normal.
• Permutable not implies normal: Combining the fact that every subgroup of the Baer norm is permutable in the whole group, but not every such subgroup is normal in the whole group, we get examples of permutable subgroups that are not normal.

Facts used

1. Omega-1 of odd-order class two p-group has prime exponent

Proof

Example of a group of prime power order

Further information: prime-cube order group:p2byp

Let $p$ be an odd prime. Let $A$ be the semidirect product of a cyclic group of order $p^{2}$ and a cyclic group of order $p$ acting nontrivially. In other words: $A$ is a group generated by $a,b$, with the relations $a^{p^{2}}=b^{p}=e$ and $ab=ba^{p+1}$:

$A:=\langle a,b\mid a^{p^{2}}=b^{p}=e,ab=ba^{p+1}\rangle$

Thus, $A$ has order $p^{3}$.

Let $B=\langle b\rangle$. $B$ is a subgroup of order $p$ in $A$. Moreover, $B$ is not normal in $A$, because it is the acting group in a semidirect product for a nontrivial action.

First, note that $\Omega _{1}(A)$ (the subgroup generated by elements of order $p$) is a subgroup of exponent $p$ (fact (1)). It cannot equal $A$, since $a^{p}\neq e$. Also, $a^{p}\in \Omega _{1}(A),b\in \Omega _{1}(A)$, so $\Omega _{1}(A)$ has order at least $p^{2}$. Hence, its order is exactly $p^{2}$, and it is elementary Abelian on $a^{p}$ and $b$.

Note that the normalizer of any subgroup of order $1,p^{2}$ or $p^{3}$ is the whole group. The normalizer of any subgroup of order $p$ must contain that subgroup, and also the center, which is generated by $a^{p}$. Hence, either the subgroup is normal, or its normalizer is a subgroup of order $p^{2}$ containing two permuting subgroups of order $p$, forcing it to be $\Omega _{1}(A)$ (again by fact (1)). Thus, every normalizer is either the whole group or $\Omega _{1}(A)$. Moreover, for the subgroup $B=\langle b\rangle$, the normalizer must be precisely $\Omega _{1}(A)$, since $B$ is not normal in $A$. Thus, the Baer norm is precisely $\Omega _{1}(A)$.

But the subgroup $B\leq \Omega _{1}(A)$ is not normal in $A$. Thus, the Baer norm is not hereditarily normal.