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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a finite group, every subgroup satisfying the first subgroup property (i.e., Abelian direct factor) must also satisfy the second subgroup property (i.e., potentially verbal subgroup). In other words, every Abelian direct factor of finite group is a potentially verbal subgroup of finite group.
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Contents 1 Statement 2 Related facts 2.1 Stronger facts 3 Facts used 4 Proof

Statement

Suppose G is a finite group and H is a direct factor of G such that H is abelian as a group. Then, there exists a group K containing G such that H is a verbal subgroup.

Related facts

Stronger facts

• Central implies potentially verbal in finite: The proof idea is the same,but it uses a central product in place of a direct product.

Facts used

1. structure theorem for finitely generated abelian groups

Proof

Given: A finite group G with direct factor H. Let L be a normal complement to H in G, so .

To prove: There exists a group K containing G such that H is a verbal subgroup of K.

Proof: If H is trivial, we can set K = G and be done. So, we can assume H is nontrivial.

1. By fact (1), we can write H as a direct product of cyclic groups, say , where each C_i is cyclic of prime power order, say .
2. For each prime p, let c(p) be a positive integer greater than the largest power of p dividing the order of L. (Note that the same p may repeat among multiple p_is).
3. Now consider the group where each D_i is cyclic of order . Treat H as the subgroup of M obtained by embedding each C_i naturally into the corresponding D_i.
4. Define , with G embedded in K using the embedding of H in M provided above.
5. Let where the product is over the primes p dividing the order of H. Then, it is clear that the set of n^th powers in K is precisely H. In particular, H is a verbal subgroup of K.