Normal subgroup contained in the hypercenter is amalgam-characteristic

Statement

Suppose $H$ is a normal subgroup of $G$ contained in the Hypercenter (?) of $G$. In other words, $H$ is contained in some member of the transfinite upper central series of $G$. Then, $H$ is an amalgam-characteristic subgroup of $G$: $H$ is characteristic in the external amalgamated free product $K:=G{*}_{H}G$.

Related facts

Similar facts

• Central implies amalgam-characteristic
• Finite normal implies amalgam-characteristic
• Periodic normal implies amalgam-characteristic

Opposite facts

• Normal not implies amalgam-characteristic
• Characteristic not implies amalgam-characteristic

Applications

• Normal subgroup contained in hypercenter is potentially characteristic
• Nilpotent implies every normal subgroup is potentially characteristic

Facts used

1. Quotient of amalgamated free product by amalgamated normal subgroup equals free product of quotient groups
2. Free product of nontrivial groups is centerless
3. Hypercenter is characteristic

Proof

Given: A group $G$, a subgroup $H$ contained in the hypercenter of $G$. $L:=G{*}_{H}G$.

To prove: $H$ is characteristic in $L$.

Proof:

1. By fact (1), $K/H\cong G/H*G/H$.
2. $K/H$ is centerless: If $H$ is proper in $G$, this follows from fact (2). If $H=G$, then $K/H$ is trivial, hence centerless.
3. $H$ is in the hypercenter of $K$: This is because $H$ is in the hypercenter of each of the factors.
4. $H$ equals the hypercenter of $K$: If $g$ is in the hypercenter of $K$, the image of $g$ via the quotient map $K\to K/H$ is in the hypercenter of $K/H$. However, since $K/H$ is centerless, we get that the image of $g$ is trivial, so $g\in H$. Thus, $H$ is the hypercenter.
5. $H$ is characteristic in $K$: This follows from the previous step and fact (3).