# 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, 2015 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, 2015 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.