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Math Help - A is a subset of B

  1. #1
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    A is a subset of B

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

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

    m\geq n

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

    <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>?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Re: A is a subset of B

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

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

    m\geq n

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

    <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 \langle X\rangle? This all follows quite quickly once you know this...
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  3. #3
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    Re: A is a subset of B

    Frankly what you have written makes little sense. If A and B are sets then " \{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?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    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 " \{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 A\subset B\subset G, and he requires to prove that \langle A\rangle\leq \langle B\rangle. However, this is clear by the very definition of \langle X\rangle!
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