# Finite size if finite number of subgroups

• Mar 7th 2006, 01:49 PM
topsquark
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
• Mar 7th 2006, 03:06 PM
ThePerfectHacker
Quote:

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 $\mathcal{G}$ be an infinite group. Then the subgroups generated by its elements (this is why it is on cyclic groups) is $=\{a_i ^n|n\in \mathbb{Z}\}$. Notice there are infinite many different $a_i$ because $\mathcal{G}$ is an infinite group. If we can show that $\not= $ for $a\not =b$ 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 $\not =$ I assumed it is true (it is definitely not true for finite groups).
• Mar 8th 2006, 12:50 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
Let $\mathcal{G}$ be an infinite group. Then the subgroups generated by its elements (this is why it is on cyclic groups) is $=\{a_i ^n|n\in \mathbb{Z}\}$. Notice there are infinite many different $a_i$ because $\mathcal{G}$ is an infinite group. If we can show that $\not= $ for $a\not =b$ then there are infinitely many subgroups (namely, one for each generating element).

Let $\mathcal{G}=(\mathbb{C}\setminus \{0\},\times)$, the group of complex numbers less 0 under multiplication.

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

RonL
• Mar 8th 2006, 05:13 AM
topsquark
Quote:

Originally Posted by CaptainBlack
Let $\mathcal{G}=(\mathbb{C}\setminus \{0\},\times)$, the group of complex numbers less 0 under multiplication.

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

RonL

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

-Dan
• Mar 8th 2006, 02:44 PM
ThePerfectHacker
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 $\mathcal{G}$ be infinite.
2)Let $S=\{|a\in\mathcal{G}\}$
3)Thus, $\bigcup x,x\in S=\mathcal{G}$.
4)Assume $S \mbox{ and }\forall a\in S$ are finite.
5)Then, by #3 $\mathcal{G}$ is finite-impossible.
6)Thus, $\mbox{ or }S$ are infinite.
7)If $S$ is infinite then the proof is complete.
8)If $ \exists a\in S$ is infinite, then $\simeq <\mathbb{Z},+>$
9)But $<\mathbb{Z},+>$ has infinitely many subgroups.
10)Thus, $$ has infinitely many subgroups.
11)If $N\leq $ then $N\leq \mathcal{G}$
12)Thus, $\mathcal{G}$ 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 $\mathcal{G}$ 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, $\mathcal{G}$ has infinitely many subgroups!
8)Up to isomorphism there is only one infinite cyclic group.
9)Namely, it multiples $n\mathbb{Z}$.
10)Because of isomorphism.
11)The relation $\leq$ 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 $a\not =e$ 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?
• Mar 9th 2006, 06:04 AM
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
• Mar 9th 2006, 01:32 PM
ThePerfectHacker
Quote:

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 :)