Results 1 to 7 of 7

Math Help - Generated subgroup

  1. #1
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6

    Generated subgroup

    Hi,

    So... I've decided to study some algebra, to see if I can handle it... lol

    I've come accross this (very basic) question, which may rather be set theory, but the original question is algebra :
    Let G be a group. Let H be a subgroup containing a subset P.
    Prove that \langle P\rangle_H=\langle P\rangle_G

    So basically, I have to prove that \bigcap_{I\in\mathcal{I}} I=\bigcap_{H\in\mathcal{H}} H

    Where \mathcal{I}=\{\text{subgroups of H containing P}\} and \mathcal{H}=\{\text{subgroups of G containing P}\}

    It is clear that \mathcal{I}\subseteq \mathcal{H}

    So it is very logical to have the equality \bigcap_{I\in\mathcal{I}} I=\bigcap_{H\in\mathcal{H}} H

    But I didn't succeed in finding a formal proof

    Thanks,
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Moo View Post
    Hi,

    So... I've decided to study some algebra, to see if I can handle it... lol

    I've come accross this (very basic) question, which may rather be set theory, but the original question is algebra :
    Let G be a group. Let H be a subgroup containing a subset P.
    Prove that \langle P\rangle_H=\langle P\rangle_G

    So basically, I have to prove that \bigcap_{I\in\mathcal{I}} I=\bigcap_{H\in\mathcal{H}} H

    Where \mathcal{I}=\{\text{subgroups of H containing P}\} and \mathcal{H}=\{\text{subgroups of G containing P}\}

    It is clear that \mathcal{I}\subseteq \mathcal{H}

    So it is very logical to have the equality \bigcap_{I\in\mathcal{I}} I=\bigcap_{H\in\mathcal{H}} H

    But I didn't succeed in finding a formal proof

    Thanks,
    I'll try not to take the set theory approach...

    Consider the following.

    If H\leq G and P\subseteq H, then as you've observed, P\subseteq H\subseteq G.

    If I understand correctly, ^n=e_H\}" alt="\left<P\right>_H=\{p\in P^n=e_H\}" />. Since H\leq G, e_H=e_G. So \left<P\right>_H=\left<P\right>_G

    This seems a bit hard to find a really formal proof since its a pretty straightforward concept...I'm not sure if this is what you're looking for.. xD
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jul 2008
    Posts
    81
    It's actually not much work to get the result, and perhaps easier to just use minimality as opposed to the intersections. In other words, I just use the fact that the subgroup generated by a set is a subgroup of any group containing the set.

    Since \langle P \rangle_H is the smallest subgroup of H containing P, it is a subgroup of G containing P, and thus by minimality of \langle P \rangle_G in G we must have \langle P \rangle_G \leq \langle P \rangle_H. Similarly, since P\subset H, by minimality in G we have \langle P \rangle_G\leq H, so by minimality in H we have \langle P \rangle_H\leq \langle P \rangle_G thus equality!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Chris L T521 View Post
    I'll try not to take the set theory approach...

    Consider the following.

    If H\leq G and P\subseteq H, then as you've observed, P\subseteq H\subseteq G.

    If I understand correctly, ^n=e_H\}" alt="\left<P\right>_H=\{p\in P^n=e_H\}" />. Since H\leq G, e_H=e_G. So \left<P\right>_H=\left<P\right>_G

    This seems a bit hard to find a really formal proof since its a pretty straightforward concept...I'm not sure if this is what you're looking for.. xD
    Hmm no it's not, sorry

    I'm technically now allowed to use generators (p), because it was further in the book

    Quote Originally Posted by siclar View Post
    It's actually not much work to get the result, and perhaps easier to just use minimality as opposed to the intersections. In other words, I just use the fact that the subgroup generated by a set is a subgroup of any group containing the set.

    Since \langle P \rangle_H is the smallest subgroup of H containing P, it is a subgroup of G containing P, and thus by minimality of \langle P \rangle_G in G we must have \langle P \rangle_G \leq \langle P \rangle_H. Similarly, since P\subset H, by minimality in G we have \langle P \rangle_G\leq H, so by minimality in H we have \langle P \rangle_H\leq \langle P \rangle_G thus equality!
    Okay, got it ! Thanks


    Now, I'd like to know if there's a formal proof for the set theory problem generated by this question
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2008
    Posts
    81
    I'm not sure what you mean by the set theory part. Is it the implication that the intersection of a subcollection of a collection of sets is equal to the intersection of the collection? That is not necessarily true. As a silly example take the sets {1,2} and {1}. Then the intersection over {1,2} is not equal to the intersection over {1,2} and {1}, you see what I mean?

    If this is not the question though, please clarify
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by siclar View Post
    I'm not sure what you mean by the set theory part. Is it the implication that the intersection of a subcollection of a collection of sets is equal to the intersection of the collection? That is not necessarily true. As a silly example take the sets {1,2} and {1}. Then the intersection over {1,2} is not equal to the intersection over {1,2} and {1}, you see what I mean?

    If this is not the question though, please clarify
    Okay, I see your point !
    My intuition was completely misleading me... no wonder I didn't succeed in proving a false thing
    I hope I understood my own problem, actually... XD

    Thanks !


    Oh, another question... How can one prove that \langle \emptyset\rangle_G=e_G ? I'm having trouble understanding this...
    Because I don't know how to interprete that a subgroup contains \emptyset
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jul 2008
    Posts
    81
    From the minimality condition, what is the smallest subgroup containing the empty set? Well, as all sets contain the empty set, this is simply the smallest subgroup which is always just the identity, the trivial subgroup!

    Oh as to "containing the empty set", this is vacuously always true. The subset statement is "If x is in A, then x is in B" for A being a subset of B. Perhaps its easier to understand from the contrapositive: "If x is not in B, then x is not in A". If A is the empty set, then the "then" part is always true hence the empty set always is a subset. I hope this clarifies things
    Last edited by siclar; July 9th 2009 at 08:36 AM. Reason: Other question in your post
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 2nd 2011, 08:07 PM
  2. subgroup generated
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: April 14th 2009, 03:48 PM
  3. Finitely generated subgroup
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 8th 2008, 08:33 PM
  4. Finitely generated subgroup implies cyclic
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 29th 2008, 12:52 AM
  5. Finitely generated subgroup of Q
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 9th 2008, 03:53 PM

Search Tags


/mathhelpforum @mathhelpforum