Central implies amalgam-characteristic

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., central subgroup) must also satisfy the second subgroup property (i.e., amalgam-characteristic subgroup)
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Statement

Statement with symbols

Suppose $H$ is a central subgroup of a group $G$. Then, $H$ is a characteristic subgroup inside the amalgam $L:=G*_{H}G$. In other words, $H$ is an amalgam-characteristic subgroup.

Definitions used

Central subgroup

Further information: Central subgroup

A subgroup $H$ of a group $G$ is termed a central subgroup if every element of $H$ commutes with every element of $G$. Equivalently, $H$ must be contained in the center of $G$.

Amalgam-characteristic subgroup

Further information: Amalgam-characteristic subgroup

A subgroup $H$ of a group $G$ is termed an amalgam-characteristic subgroup if $H$ is a characteristic subgroup inside the amalgam $L:=G*_{H}G$.

Related facts

Similar facts

• Normal subgroup contained in hypercenter is 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

• Central implies potentially characteristic
• Abelian implies every subgroup is potentially characteristic

Facts used

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

Proof

Given: A group $G$, a central subgroup $H$. $L:=G*_{H}G$.

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

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$L/H\cong G/H*G/H$.	Facts (1), (2)	$H$ is central in $G$		Fact-direct
2	$L/H$ is centerless.	Fact (3)		Step (1)	[SHOW MORE] If $H$ is a proper subgroup of $G$, then by Fact (2), $L/H$ is centerless. If $H=G$, then $L/H$ is trivial and hence centerless.
3	$H$ is in the center of $L$.		$H$ is central in $G$		[SHOW MORE] $H$ is central in $G$, hence it commutes with elements in both copies of $G$, Since the two copies of $G$ generate $L$, $H$ commutes with all of $L$ and is hence in the center of $L$.
4	$H$ equals the center of $L$.			Steps (2), (3)	[SHOW MORE] Step (3) already showed that $H$ is in the center, so we only need to show that if $z$ is central in $L$, then $z\in H$. Let's do this. Consider the image of $z$ in $L/H$. Since $z$ is central in $L$, this image is central in $L/H$. By Step (2), however, $L/H$ is centerless, forcing this image to be the identity element, thus forcing $z\in H$ as required.
5	$H$ is characteristic in $G$.	Fact (4)		Step (4)	Step-fact combination direct

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