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

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

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

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

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



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

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

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

    Therefore $\displaystyle G$ cannot be the union of two proper subgroups.


    Thank you.
    Hjörtur
    Last edited by hjortur; Jun 4th 2010 at 05:58 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 $\displaystyle (M,p,1)$ and let $\displaystyle m\in M$. Define a new product $\displaystyle p_m$ in $\displaystyle M$ by $\displaystyle p_m(a,b)=amb$.
    Show that this defines a semigroup. Under what conditions on $\displaystyle m$ do we have a unit relative to $\displaystyle p_m$?

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

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

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

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



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

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

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

    Therefore $\displaystyle G$ 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 $\displaystyle (M, p, 1)$ was a monoid with multiplication $\displaystyle p$.
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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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