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Math Help - Moniod with a unit

  1. #1
    Senior Member sfspitfire23's Avatar
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    Moniod with a unit

    Hello!

    If M is a finite monoid and au=bu in M implies that a=b, show that u is a unit.

    with hints: If M=\{a_1,\cdots a_n\} show that a_1u,\cdots , a_nu are distinct.


    Attempt:

    If M=\{a_1,\cdots a_n\} then by definition it has an identity element e. Then, because M is finite, there will be an element which will map to the identity of M. Each a_1u\cdots a_nu will be distinct as for every a_n, u will map it back to itself. Thus, au=bu which, because u will map a to itself a(e)=b(e) and a=b(e) so u must be a unit.



    Thanks guys!
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  2. #2
    Senior Member
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    Hi

    What you have to prove is: \exists b\in M\ \text{s.t.}\ bu=e.

    To show the hint, just use that au=bu\Rightarrow a=b, (for instance prove it by contraposition).

    Assume you've proved it. Show that M=\{a_1u,...,a_nu\} (using a property of finite sets)

    Therefore e\in M means e\in \{a_1u,...,a_nu\}. Can you conclude?
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  3. #3
    Senior Member sfspitfire23's Avatar
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    would it suffice to create a cayley table with elements 1,a,b and show that a and b both have an inverse? ugh, where would i go from there?


    thanks, really appreciate it
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  4. #4
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    I don't really understand, why just e,a,b in a Cayley table. ab can be different from e,a or b.

    What the proof says is : the condition you have implies that the map M\rightarrow M: a\mapsto au is injective and then surjective (because it's an injective map between sets of same finite cardinality) therefore e has an antecedant, which is some b\in M such that bu=e, and we're done.
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