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Math Help - 2 Basic algebra questions

  1. #1
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    2 Basic algebra questions

    I am studying by my self Basic Algebra by Nathan Jacobson, and I am currently doing the exercises from chapter 1.
    I would be very thankful if someone would look over my answers and
    tell me if my argument is sound.


    1. Let (M,p,1) be a monoid with multiplication p, and let m\in M. Define a new product p_m in M by p_m(a,b)=amb.
    Show that this defines a semigroup. Under what conditions on m do we have a unit relative to p_m?

    For this to be a semigroup then we have to show that p_m is associative. Let * denote p_m.
    Let a,b,c \in M, then

    (a*b)*c=(amb)mc=am(bmc)=a*(b*c), because composition under p is associative, so p_m is associative so this is a semigroup.

    Now we do have a unit relative to p_m if and only if m has an inverse under p. Then the unit relative to p_m is m^{-1}.
    That is

    a*m^{-1}=amm^{-1}=a1=a and m^{-1}*a=m^{-1}ma=1a=a for all a\in M



    2. Show that G cannot be the union of two proper subgroups.

    Lets assume that G is the union of proper subgroups A,B. Since they are proper there exist elements a\in A, b\in B such that a\notin B and b\notin A.
    Now since G is a group then it is closed under composition, that is ab\in G. Then ab\in A or ab\in B.

    Since A is a subgroup and a\in A then a^{-1} \in A and A is closed under composition. Then a^{-1}\cdot ab=b\in A but that is a contradiction. In the same way
    ab\in B leads to a\in B which is also a contradiction.

    Therefore G cannot be the union of two proper subgroups.


    Thank you.
    Hjörtur
    Last edited by hjortur; June 4th 2010 at 06:58 AM.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by hjortur View Post
    I am studying by my self Basic Algebra by Nathan Jacobson, and I am currently doing the exercises from chapter 1.
    I would be very thankful if someone would look over my answers and
    tell me if my argument is sound.


    1. Let (M,p,1) and let m\in M. Define a new product p_m in M by p_m(a,b)=amb.
    Show that this defines a semigroup. Under what conditions on m do we have a unit relative to p_m?

    For this to be a semigroup then we have to show that p_m is associative. Let * denote p_m.
    Let a,b,c \in M, then

    (a*b)*c=(amb)mc=am(bmc)=a*(b*c), because composition under p is associative, so p_m is associative so this is a semigroup.

    Now we do have a unit relative to p_m if and only if m has an inverse under p. Then the unit relative to p_m is m^{-1}.
    That is

    a*m^{-1}=amm^{-1}=a1=a and m^{-1}*a=m^{-1}ma=1a=a for all a\in M



    2. Show that G cannot be the union of two proper subgroups.

    Lets assume that G is the union of proper subgroups A,B. Since they are proper there exist elements a\in A, b\in B such that a\notin B and b\notin A.
    Now since G is a group then it is closed under composition, that is ab\in G. Then ab\in A or ab\in B.

    Since A is a subgroup and a\in A then a^{-1} \in A and A is closed under composition. Then a^{-1}\cdot ab=b\in A but that is a contradiction. In the same way
    ab\in B leads to a\in B which is also a contradiction.

    Therefore G cannot be the union of two proper subgroups.


    Thank you.
    Hjörtur
    That seems good to me, although it would have helped if you mentioned that (M, p, 1) was a monoid with multiplication p.
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  3. #3
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    Thank you, I edited the post to fix this. I dont know how I could forget to type the latter half of the sentence.
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