Characteristic not implies elementarily characteristic

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) need not satisfy the second subgroup property (i.e., elementarily characteristic subgroup)
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Statement

It is possible to have a characteristic subgroup $H$ of a group $G$ that is not an elementarily characteristic subgroup of $G$, i.e., there exists a subgroup $K$ of $G$ such that the first-order theories of the group-subgroup pairs $(G,H)$ and $(G,K)$ are elementarily equivalent.

Proof

Suppose $T_{1},T_{2}$ are two infinite sets with $|T_{1}|>|T_{2}|$. Let $G$ be the direct product $G_{1}\times G_{2}$ where $G_{1}$ is the finitary alternating group on $T_{1}$ and $G_{2}$ is the finitary alternating group on $T_{2}$. Let $H$ be the subgroup of $G$ that is the first direct factor ($G_{1}\times \{e\}$) and $K$ be the subgroup of $G$ that is the second direct factor ($\{e\}\times G_{2}$).

• $H$ is characteristic (in fact, homomorph-containing) in $G$: By cardinality considerations and the fact that $H$ is simple, any homomorphic image of $H$ in $G_{2}$ is trivial. Therefore, the image of $H$ under any homomorphism of $G$ is contained in $H$. Therefore, $H$ is characteristic in $G$.
• The first-order theories of the group-subgroup pairs $(G,H)$ and $(G,K)$ are elementarily equivalent, so $H$ is not elementarily characteristic in $G$: This can be seen from the idea that first-order statements reference only finite subsets, and as far as finite subsets are concerned, $G_{1}$ and $G_{2}$ look the same.