2
$\begingroup$

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!)

$\endgroup$
2
  • $\begingroup$ 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?" $\endgroup$
    – Fizz
    Apr 12, 2015 at 1:48
  • 1
    $\begingroup$ 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. $\endgroup$
    – Fizz
    Apr 12, 2015 at 5:29

2 Answers 2

Reset to default
8
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ I liked this answer because it helped me realize we can study the properties of an infinite set by studying the finite components that make it up. Thanks. $\endgroup$
    – millsmess
    Apr 11, 2015 at 23:17
  • $\begingroup$ @millsmess: When you go on to other mathematical structures, including those in logic, linear algebra, field theory and topology, you will find that in general iteratively defined infinite structures almost always have properties that arise from the properties of the finite subsets. For example, in any polynomial ring with infinitely many variables, any polynomial still has finitely many variables, and so usually can be first studied as an element of a polynomial ring with finitely many variables before we lift the properties back to the original ring. $\endgroup$
    – user21820
    Apr 12, 2015 at 6:46
3
$\begingroup$

It is easier than it looks. All you need to do is see that the conditions for being a subgroup are satisfied. Here's a hint: for every pair of elements in the union there is some subgroup in the sequence containing both of them.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .