# How to show that the union of an infinite sequence of subgroups is a subgroup?

I'm self-studying from Algebra: Chapter 0 by Aluffi. I was working on Exercise 6.6, and I can do the first part (which I've included for context), but I'm having trouble with the second part.

Here is Exercise 6.6:

Prove that the union of a family of subgroups of a group G is not necessarily a subgroup of G. In fact:

• Let $H$, $H'$ be subgroups of a group $G$. Prove that $H \cup H'$ is a subgroup of G only if $H \subseteq H'$ or $H' \subseteq H$.

• On the other hand, let $H_0 \subseteq H_1 \subseteq H_2 \subseteq ...$ be subgroups of a group $G$. Prove that $\cup_{i\geq0}H_i$ is a subgroup of $G$.

The second part is easy in the case that the sequence of subgroups is finite. Say that there is a last subgroup in the sequence - call it $H_{last}$. Then $\cup_{i\geq0}H_i$ is just $H_{last}$, which we know is a subgroup of G.

However, I'm really not sure how to approach the second part if the sequence is infinite. Do we have to do something analogous to taking a limit in calculus?

(I'm an engineering major and haven't taken any analysis or abstract algebra classes, so if you could keep that in mind when answering that would be great!)

• You may want to review the def of infinite union. Halmos' Naive Set Theory is pretty detailed on this ("Section 9: Families"). This is addressing your sub-question "Do we have to do something analogous to taking a limit in calculus?" – Fizz Apr 12 '15 at 1:48
• You should also try to come up with a concrete example where $H_0 \subseteq H_1 \subseteq H_2 \subseteq ...$ is an infinite chain of subgroups. – Fizz Apr 12 '15 at 5:29

Let $H=\bigcup_{i\geq 0}H_i$. The hardest property is probably checking $H$ is closed under composition. Let $a,b\in H$. Then $a\in H_i$ and $b\in H_j$ for some $i$ and $j$. Without loss of generality, we can assume $i\leq j$, so $H_i\subseteq H_j$. Thus $a,b\in H_j$, so $ab\in H_j$, since $H_j$ is a subgroup, so $ab\in H$.
The other properties are not bad. Just realize that if $a\in H$, then $a\in H_i$ for some $i$, and use the fact that $H_i$ is a subgroup itself.