Results 1 to 4 of 4

Thread: A is a subset of B

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    A is a subset of B

    Prove that if A is a subset of B, then $\displaystyle <A>\leq <B>$.

    Let $\displaystyle |A|=n \ \text{and} \ |B|=m$

    $\displaystyle m\geq n$

    $\displaystyle <A>=\{e,a,a^2,\cdots, a^{n-1}\}$

    $\displaystyle <B>=\{e,b,b^2,\cdots, b^{m-1}\}$

    Can I just since |A| less than or equal to |B|, <A> is a subgroup of <B>?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176

    Re: A is a subset of B

    Quote Originally Posted by dwsmith View Post
    Prove that if A is a subset of B, then $\displaystyle <A>\leq <B>$.

    Let $\displaystyle |A|=n \ \text{and} \ |B|=m$

    $\displaystyle m\geq n$

    $\displaystyle <A>=\{e,a,a^2,\cdots, a^{n-1}\}$

    $\displaystyle <B>=\{e,b,b^2,\cdots, b^{m-1}\}$

    Can I just since |A| less than or equal to |B|, <A> is a subgroup of <B>?
    What is the written down-definition for $\displaystyle \langle X\rangle$? This all follows quite quickly once you know this...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,729
    Thanks
    3010

    Re: A is a subset of B

    Frankly what you have written makes little sense. If A and B are sets then "$\displaystyle \{e, a, a^2, ..., a^{n-1}\}$" is meaningless because there is no operation in a set. Apparently you are talking about groups, not sets. But even then you have to define <A>. You appear to mean a sub-group generated by a single element of the group but which element? Or are you assuming from the start that A and B are themselves generated by a single element?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176

    Re: A is a subset of B

    Quote Originally Posted by HallsofIvy View Post
    Frankly what you have written makes little sense. If A and B are sets then "$\displaystyle \{e, a, a^2, ..., a^{n-1}\}$" is meaningless because there is no operation in a set. Apparently you are talking about groups, not sets. But even then you have to define <A>. You appear to mean a sub-group generated by a single element of the group but which element? Or are you assuming from the start that A and B are themselves generated by a single element?
    What he means is $\displaystyle A\subset B\subset G$, and he requires to prove that $\displaystyle \langle A\rangle\leq \langle B\rangle$. However, this is clear by the very definition of $\displaystyle \langle X\rangle$!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Proof that if A is a subset of P(A), P(A) is a subset of P(P(A))
    Posted in the Discrete Math Forum
    Replies: 12
    Last Post: May 23rd 2011, 04:26 PM
  2. Proof on Openness of a Subset and a Function of This Subset
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Oct 24th 2010, 09:04 PM
  3. subset help
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Sep 23rd 2009, 09:08 PM
  4. subset U subset is NOT subspace of Superset?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Sep 20th 2009, 04:17 PM
  5. Is X a subset of X * Y
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Oct 1st 2008, 05:53 PM

/mathhelpforum @mathhelpforum