Let $\left\{ H_i | i \in I\right\}$ be a family of subgroups of G. State and prove a condition which makes $\cup_{i \in I} H_{i}$ a subgroup of $G$

Question

Let $G$ be a group and $\left\{ H_i | i \in I\right\}$ be a family of subgroups of G. State and prove a condition which makes $\cup_{i \in I} H_{i}$ a subgroup of $G$, that is that $\cup _{i \in I} H_{i} = \left< \cup_{i \in I} H_{i} \right>$.

To show that $\cup _{i \in I} H_{i} \subseteq \left< \cup _{i \in I} H_{i} \right>$, it seems that I have to show that $\cup _{i \in I} H_{i}$ is a subgroup of G. Since $\left< \cup _{i \in I} H_{i} \right>$ is a collection of subgroups of G that contain $\cup _{i \in I} H_{i}$ as a subset, I would have the first inclusion proven and quite possibly get the second inclusion for free.

However, I am having a hard time on how to prove $\cup _{i \in I} H_{i}$ is a subgroup of $G$. The only way this could happen is the collection of subgroups $\{ H_{i} | i \in I\}$ is a chain.

Is there another way to show that $\cup _{i \in I} H_{i}$ is a subgroup of $G$?

Am I thinking about this the right way?

Answer 1

As stated, it seems like the condition is of your choosing. You are on the right track when thinking about a collection of 'chained' groups. I would advise you to try and prove this for a countable collection $G_1 \subset G_2 \dots$. This could lead you to stronger results which only use the essentials of the previous proof.

A sketch of a possible generalization is below,

Suppose that $(H_i)_{i \in I}$ is such that for each $g \in H_i, g' \in H_j$ there exists $k \in I$ with $g,g' \in H_k$. Then, the set $H = \bigcup_{i \in I} H_i$ is a subgroup of $G$: clearly we have that $1 \in G$; and if $g \in H_i \subset H$, then $g^{-1} \in H_i \subset H$. So far we have used no assumptions. Now, take $g,g' \in H$. Hence there exist $i,j \in I$ with $g \in H_i$ and $g \in H_j$. By hypothesis we have $k \in I$ with $g,g' \in H_k$ and so $gg' \in H_k \subset H \ \square$.