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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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Contents 1 Statement 1.1 Statement with symbols 2 Definitions used 2.1 Central subgroup 2.2 Amalgam-characteristic subgroup 3 Related facts 3.1 Similar facts 3.2 Opposite facts 3.3 Applications 4 Facts used 5 Proof

Statement

Statement with symbols

Suppose H is a central subgroup of a group G. Then, H is a characteristic subgroup inside the amalgam K: = G * _HG. 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 * _HG.

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

Proof

Given: A group G, a central subgroup H. L: = G * _HG.

To prove: H is characteristic in L.

Proof:

1. By fact (1), .
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 center of K: This is because H is in the center of each of the factors.
4. H equals the center of K: If g is in the center of K, the image of g via the quotient map is in the center of K / H. However, since K / H is centerless, we get that the image of g is trivial, so . Thus, H is the center.
5. H is characteristic in K: This follows from the previous step and fact (3).