Finite normal 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., finite normal subgroup) must also satisfy the second subgroup property (i.e., amalgam-characteristic subgroup)
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This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., finite group) must also satisfy the second group property (i.e., always amalgam-characteristic group)
View all group property implications | View all group property non-implications
Get more facts about finite group|Get more facts about always amalgam-characteristic group

Statement

Verbal statement

Any finite normal subgroup (i.e., a normal subgroup that is finite as a group) is an amalgam-characteristic subgroup.

Equivalently, a finite group is an always amalgam-characteristic group: it is an amalgam-characteristic subgroup in any group in which it is a normal subgroup.

Statement with symbols

Suppose $H$ is a finite normal subgroup of a group $G$. Then, $H$ is a characteristic subgroup inside the amalgam $L:=G{*}_{H}G$.

Related facts

Stronger facts

• Finite normal implies amalgam-normal-subhomomorph-containing

Similar facts

• Central implies amalgam-characteristic
• Periodic normal implies amalgam-characteristic
• Normal subgroup contained in hypercenter is amalgam-characteristic

Opposite facts

• Normal not implies amalgam-characteristic
• Characteristic not implies amalgam-characteristic
• Direct factor not implies amalgam-characteristic
• Cocentral not implies amalgam-characteristic

Applications

• Finite normal implies potentially characteristic
• Finite normal implies image-potentially characteristic

Facts used

1. Normality is upper join-closed
2. Quotient of amalgamated free product by amalgamated normal subgroup equals free product of quotient groups
3. Free product of nontrivial groups has no nontrivial finite normal subgroup
4. Normality satisfies image condition

Proof

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

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

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$H$ is a finite normal subgroup of $L$.	Fact (1)	$H$ is finite and normal in $G$		Since $H$ is normal in both copies of $G$, it is normal in $L$. Also, $H$ is finite.
2	$L/H\cong G/H*G/H$.	Fact (2)	$H$ is normal in $G$		Fact-direct
3	$L/H$ has no nontrivial finite normal subgroup.	Fact (3)		Step (2)	[SHOW MORE] If $H$ is a proper subgroup of $G$, then by Fact (3), $L/H$ has no nontrivial finite normal subgroup. If $H=G$, then $L/H$ is trivial and hence has no nontrivial finite normal subgroup.
4	$H$ is the unique largest finite normal subgroup of $L$.	Fact (4)		Steps (1), (3)	[SHOW MORE] Step (1) established that $H$ is finite and normal, so we only need to show that every finite normal subgroup of $L$ is contained in $H$. Suppose $M$ is a finite normal subgroup of $L$. Then, by fact (4), the image of $M$ in the quotient map $L\to L/H$ is a normal subgroup of $L/H$. Also, since $M$ is finite, its image is finite. In Step (3), we concluded that $L/H$ has no nontrivial finite normal subgroup. Thus, the image of $M$ is trivial, so $M\le H$.
5	$H$ is characteristic in $L$.			Step (4)	[SHOW MORE] The condition of being the unique largest finite normal subgroup is invariant under automorphisms, so by Step (4), $H$ must be characteristic in $L$.

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