# Thread: Finite size if finite number of subgroups

1. ## 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

2. 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 $\displaystyle \mathcal{G}$ be an infinite group. Then the subgroups generated by its elements (this is why it is on cyclic groups) is $\displaystyle <a_i>=\{a_i ^n|n\in \mathbb{Z}\}$. Notice there are infinite many different $\displaystyle a_i$ because $\displaystyle \mathcal{G}$ is an infinite group. If we can show that $\displaystyle <a>\not= <b>$ for $\displaystyle a\not =b$ 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 $\displaystyle <a>\not =<b>$ I assumed it is true (it is definitely not true for finite groups).

3. Originally Posted by ThePerfectHacker
Let $\displaystyle \mathcal{G}$ be an infinite group. Then the subgroups generated by its elements (this is why it is on cyclic groups) is $\displaystyle <a_i>=\{a_i ^n|n\in \mathbb{Z}\}$. Notice there are infinite many different $\displaystyle a_i$ because $\displaystyle \mathcal{G}$ is an infinite group. If we can show that $\displaystyle <a>\not= <b>$ for $\displaystyle a\not =b$ then there are infinitely many subgroups (namely, one for each generating element).
Let $\displaystyle \mathcal{G}=(\mathbb{C}\setminus \{0\},\times)$, the group of complex numbers less 0 under multiplication.

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

RonL

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

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

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

-Dan

5. 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?

6. 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

7. 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