Simplicity is directed union-closed

This article gives the statement, and possibly proof, of a group property (i.e., simple group) satisfying a group metaproperty (i.e., directed union-closed group property)
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Statement

Verbal statement

A directed union of simple subgroups is simple.

Statement with symbols

Suppose $G$ is a group, $I$ is a nonempty directed set, and $H_{i},i\in I$ is a collection of subgroups such that $i<j\implies H_{i}\leq H_{j}$. Then, if all the $H_{i}$s are simple, so is their union.

Definitions used

Directed set

A directed set is a partially ordered set $(I,\leq )$ such that if $i,j\in I$, there exists $k\in I$, such that $i\leq k,j\leq k$.

Simple group

A nontrivial group is simple if it has no proper nontrivial normal subgroup.

Facts used

1. Directed union of subgroups is subgroup
2. Normality satisfies transfer condition: The intersection of a normal subgroup with another subgroup is normal in that subgroup.

Proof

Given: A group $G$, a collection of simple subgroups $H_{i}$ indexed by a directed set $I$, such that $i<j\implies H_{i}\leq H_{j}$.

To prove: The union of $H_{i}$ is a simple subgroup.

Proof: Suppose $H$ is the union of the $H_{i}$s. By fact (1), $H$ is a subgroup, so it suffices to show that $H$ is simple. We do this by showing that any normal subgroup $N$ of $H$ is either trivial or equal to $H$.

For each $i$, consider $N\cap H_{i}$. By fact (2), this is a normal subgroup of $H_{i}$, hence is either trivial or equals the whole of $H_{i}$. Suppose $N\cap H_{i}=H_{i}$. Then, $N$ is a normal subgroup of $H$ containing $H_{i}$.

Now, consider any subgroup $H_{j},j\neq i$. By the definition of directed set, there exists $k\in I$ such that $i,j<k$. So, $H_{k}$ contains both $H_{i}$ and $H_{j}$. Thus, the intersection $N\cap H_{k}$ is nontrivial (since the intersection contains $H_{i}$). By the simplicity of $H_{k}$, and fact (2) again, $N\cap H_{k}=H_{k}$, so $H_{k}\leq N$. In particular, $H_{j}\leq N$. Thus, every $H_{j},j\in I$, is contained in $N$, Thus, $N=H$.

Thus, if $N\cap H_{i}=H_{i}$ for any $i\in I$, $N=H$. This leaves the case that $N\cap H_{i}$ is trivial for every $i\in I$, forcing $N\cap (\bigcup _{i})H_{i}$ to be trivial, and thus forcing $N\cap H=N$ to be trivial.