Results 1 to 7 of 7

Math Help - Finite size if finite number of subgroups

  1. #1
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,054
    Thanks
    368
    Awards
    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    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>=\{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 <a>\not= <b> 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 <a>\not =<b> I assumed it is true (it is definitely not true for finite groups).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    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>=\{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 <a>\not= <b> 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 <a>=<b>, but a \ne b.

    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,054
    Thanks
    368
    Awards
    1
    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 <a>=<b>, 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
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    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.
    --------
    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>|a\in\mathcal{G}\}
    3)Thus, \bigcup x,x\in S=\mathcal{G}.
    4)Assume S \mbox{ and }<a>\forall a\in S are finite.
    5)Then, by #3 \mathcal{G} is finite-impossible.
    6)Thus, <a>\mbox{ or }S are infinite.
    7)If S is infinite then the proof is complete.
    8)If <a> \exists a\in S is infinite, then <a>\simeq <\mathbb{Z},+>
    9)But <\mathbb{Z},+> has infinitely many subgroups.
    10)Thus, <a> has infinitely many subgroups.
    11)If N\leq <a> 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.
    --------------------
    Anyways I had fun solving this problem. What I was first tried to show that if a\not =e then, <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?
    Last edited by ThePerfectHacker; March 8th 2006 at 07:18 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,054
    Thanks
    368
    Awards
    1
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. When the size of two finite sets is equal
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: December 22nd 2011, 02:57 AM
  2. [SOLVED] Cyclic Subgroups of a finite group G
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 10th 2011, 10:05 AM
  3. Poincare's theorem about subgroups of finite index
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 26th 2009, 11:19 AM
  4. subgroups of finite cyclic groups
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: August 30th 2009, 05:47 PM
  5. show has no proper subgroups of finite index
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 26th 2008, 07:36 PM

Search Tags


/mathhelpforum @mathhelpforum