Complete and potentially fully invariant implies homomorph-containing

Statement

Suppose $H$ is a subgroup of $G$ such that $H$ is a complete group. Suppose further that $H$ is a potentially fully invariant subgroup of $G$, i.e., there is a group $K$ containing $G$ such that $H$ is a fully invariant subgroup of $K$. Then, $H$ is a homomorph-containing subgroup of $G$: it contains every homomorphic image of itself in $G$.

Related facts

• Fully normalized and potentially fully invariant implies centralizer-annihilating endomomorphism-invariant
• Normal not implies potentially fully invariant

Facts used

1. Fully invariant implies normal
2. Equivalence of definitions of complete direct factor: A complete subgroup is a normal subgroup if and only if it is a direct factor.
3. Equivalence of definitions of fully invariant direct factor: For a direct factor, being fully invariant is equivalent to being a homomorph-containing subgroup.
4. Homomorph-containment satisfies intermediate subgroup condition

Proof

Given: A group $G$, a complete subgroup $H$. A group $K$ containing $G$ such that $H$ is fully invariant in $K$.

To prove: $H$ is a homomorph-containing subgroup of $G$.

Proof:

1. $H$ is a normal subgroup of $K$: This follows from fact (1).
2. $H$ is a direct factor of $K$: This follows from the previous step and fact (2).
3. $H$ is a homomorph-containing subgroup of $K$: This follows from the previous step and fact (3).
4. $H$ is a homomorph-containing subgroup of $G$: This follows from the previous step and fact (4).