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

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- Mar 7th 2006, 12:49 PMtopsquarkFinite 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 - Mar 7th 2006, 02:06 PMThePerfectHackerQuote:

Originally Posted by**topsquark**

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 $\displaystyle <a>\not =<b>$ I assumed it is true (it is definitely not true for finite groups). - Mar 7th 2006, 11:50 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

Let $\displaystyle a=i$ and $\displaystyle b=-i$, then $\displaystyle <a>=<b>$, but $\displaystyle a \ne b$.

RonL - Mar 8th 2006, 04:13 AMtopsquarkQuote:

Originally Posted by**CaptainBlack**

-Dan - Mar 8th 2006, 01:44 PMThePerfectHacker
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 $\displaystyle \mathcal{G}$ be infinite.

2)Let $\displaystyle S=\{<a>|a\in\mathcal{G}\}$

3)Thus, $\displaystyle \bigcup x,x\in S=\mathcal{G}$.

4)Assume $\displaystyle S \mbox{ and }<a>\forall a\in S$ are finite.

5)Then, by #3 $\displaystyle \mathcal{G}$ is finite-impossible.

6)Thus, $\displaystyle <a>\mbox{ or }S$ are infinite.

7)If $\displaystyle S$ is infinite then the proof is complete.

8)If $\displaystyle <a> \exists a\in S$ is infinite, then $\displaystyle <a>\simeq <\mathbb{Z},+>$

9)But $\displaystyle <\mathbb{Z},+>$ has infinitely many subgroups.

10)Thus, $\displaystyle <a>$ has infinitely many subgroups.

11)If $\displaystyle N\leq <a>$ then $\displaystyle N\leq \mathcal{G}$

12)Thus, $\displaystyle \mathcal{G}$ 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 $\displaystyle \mathcal{G}$ 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, $\displaystyle \mathcal{G}$ has infinitely many subgroups!

8)Up to isomorphism there is only one infinite cyclic group.

9)Namely, it multiples $\displaystyle n\mathbb{Z}$.

10)Because of isomorphism.

11)The relation $\displaystyle \leq$ 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 $\displaystyle a\not =e$ then, $\displaystyle <a>$ 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? - Mar 9th 2006, 05:04 AMtopsquark
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 - Mar 9th 2006, 12:32 PMThePerfectHackerQuote:

Originally Posted by**topsquark**