Thread: Finite size if finite number of subgroups

Here's another problem that likely has a simple answer...

"Show that a group G has finite cardinality if it has a finite number of subgroups."

I have no real idea where to start except that the question is in the section on cyclic groups.

-Dan

Originally Posted by topsquark

Here's another problem that likely has a simple answer...

"Show that a group G has finite cardinality if it has a finite number of subgroups."

I have no real idea where to start except that the question is in the section on cyclic groups.

-Dan

Let be an infinite group. Then the subgroups generated by its elements (this is why it is on cyclic groups) is . Notice there are infinite many different because is an infinite group. If we can show that for then there are infinitely many subgroups (namely, one for each generating element).
Now we know the following:
If an infinite group then infinite subgroups
Take contrapositive:
If finite subgroups then finite group.
Q.E.D.

I did not prove that I assumed it is true (it is definitely not true for finite groups).

Originally Posted by ThePerfectHacker

Let be an infinite group. Then the subgroups generated by its elements (this is why it is on cyclic groups) is . Notice there are infinite many different because is an infinite group. If we can show that for then there are infinitely many subgroups (namely, one for each generating element).

Let , the group of complex numbers less 0 under multiplication.

Let and , then , but .

RonL

Originally Posted by CaptainBlack

Let , the group of complex numbers less 0 under multiplication.

Let and , then , but .

RonL

Hmmm...The plot thickens! I'll try to take it from here and see what happens. Thanks to both of you!

-Dan

CaptainBlack and topsquark I apologize. I indeed made a mistake in my proof, I realized that when I was falling asleep. During, school I was thinking of a proof and I found one.
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I will prove the logically equivalent statement. That an infinite group has infinitely many subgroups (then proceede with the contrapositive).

1)Let be infinite.
2)Let
3)Thus, .
4)Assume are finite.
5)Then, by #3 is finite-impossible.
6)Thus, are infinite.
7)If is infinite then the proof is complete.
8)If is infinite, then
9)But has infinitely many subgroups.
10)Thus, has infinitely many subgroups.
11)If then
12)Thus, has infinitely many subgroups.

To help understand the proof here are some notes on each step:
1)The infinite group (any).
2)The set of all subgroups generated by an element.
3)Because all the elements of appear in this set.
4)Assumption.
5)The uninon of finitely many finite sets is a finite set.
6)De' Morgan's Laws of negation.
7)Thus, has infinitely many subgroups!
8)Up to isomorphism there is only one infinite cyclic group.
9)Namely, it multiples .
10)Because of isomorphism.
11)The relation is subgroup relation.
12)Q.E.D.
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Anyways I had fun solving this problem. What I was first tried to show that if then, in an infinite group. It made sense but I was unable to prove it. Later on I realized that was not necessary. Would you know the answer to my question?

Well, I like to think I'm clever, but I'm sure I wouldn't have gone quite that route. It's simple enough once it's laid out, but I never would have gotten the process without seeing it first.

Thank you!

-Dan

Originally Posted by topsquark

Well, I like to think I'm clever, but I'm sure I wouldn't have gone quite that route. It's simple enough once it's laid out, but I never would have gotten the process without seeing it first.

Thank you!

-Dan

Thank you