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 $i<j⟹H_i\le H_j$. 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 $(I,\le )$ such that if $i,j\in I$, there exists $k\in I$, such that $i\le k,j\le k$.

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 $i<j⟹H_i\le H_j$.

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 $H_j,j\ne i$. 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 $H_k\le N$. In particular, $H_j\le N$. 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.