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Thread: Two binary operations on a group

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    Two binary operations on a group

    This is more of a lark than anything else.

    I've been working with rings and I had a thought. Rings have two binary operations on the elements. What if we have a multiplicative group and wanted to impose a ring structure on it? My example is say we have $\displaystyle D_6 : \{r^3 = s^2 = 1,~rs = s r^{-1} \}$. Is there a process to write an additive binary operation for this? (I doubt it would have a geometric meaning in this case.) I can construct one but there are a number of arbitrary choices to make. Is there a "standard" procedure to do this in a systematic way (ie. get rid of the arbitrary choices) that would make the system a ring? I suspect that the different addition tables I can write are the same up to a relabeling of the elements and I haven't actually checked to see if the addition tables are associative. I don't know how to check to see if the elements are a group under the addition operation in general.

    Just a side thought.

    -Dan
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    Re: Two binary operations on a group

    Quote Originally Posted by topsquark View Post
    This is more of a lark than anything else.

    I've been working with rings and I had a thought. Rings have two binary operations on the elements. What if we have a multiplicative group and wanted to impose a ring structure on it? My example is say we have $\displaystyle D_6 : \{r^3 = s^2 = 1,~rs = s r^{-1} \}$. Is there a process to write an additive binary operation for this? (I doubt it would have a geometric meaning in this case.) I can construct one but there are a number of arbitrary choices to make. Is there a "standard" procedure to do this in a systematic way (ie. get rid of the arbitrary choices) that would make the system a ring? I suspect that the different addition tables I can write are the same up to a relabeling of the elements and I haven't actually checked to see if the addition tables are associative. I don't know how to check to see if the elements are a group under the addition operation in general.
    Groups have one operation.
    Rings have two operations.
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    Re: Two binary operations on a group

    Quote Originally Posted by Plato View Post
    Groups have one operation.
    Rings have two operations.
    I'm starting with what I'm treating as a multiplicative group and defining an addition operation to the set elements.

    -Dan
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    Re: Two binary operations on a group

    If you wanted, you could define $\forall r,s \in D_6, r+s=rs$.
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    Re: Two binary operations on a group

    So make the "addition" the same as the "multiplication"? But then multiplication would not "distribute" over addition. If a+ b= ab then a(b+ c)= abc while ab+ ac= abac
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    Re: Two binary operations on a group

    Quote Originally Posted by HallsofIvy View Post
    So make the "addition" the same as the "multiplication"? But then multiplication would not "distribute" over addition. If a+ b= ab then a(b+ c)= abc while ab+ ac= abac
    Good point. My bad.
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    Re: Two binary operations on a group

    @SlipEternal: Good try, though. If I'm going to make a ring the addition operation has to be defined to be Abelian. I have to get going right now but I'll be back later to show an example.

    -Dan
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    Re: Two binary operations on a group

    Here's an example. Say we have the Klein 4-group, $\displaystyle V_4$, which I'm calling my multiplication. I need to find an Abelian additive group to pair with it such that combining the two operations gives me a ring. The first thing to do is assign a new element, an additive identity, 0. This is a negative example because the only possible group with five elements is $\displaystyle Z_5$ and the distributive laws fail. For example, $\displaystyle 2 \cdot (2 + 1) = 2 \cdot 2 + 2 \cdot 1 = 1 + 2 = 3$ whereas $\displaystyle 2 \cdot ( 2 + 1) = 2 \cdot 3 = 4 \neq 3$. It's the property that $\displaystyle a^2 = 1 ~ \forall ~ a \in V_4$ that causes the problem.

    That doesn't mean, though, that there is no way to add an additive operation to a multiplicative group for at least some groups that will give a ring structure. Some examples work: Let $\displaystyle Z_p ^{\times} = Z_p - \{0\}$, where p is a prime, be the multiplication and $\displaystyle Z_p$ the additive group. I believe this gives a ring. (If I'm not mistaken it's actually a field.) My question is if there is a way to start with a group defined as multiplication and derive an additive group to go along with it such that the combined operations form a ring?

    -Dan
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    Re: Two binary operations on a group

    How about

    $$\begin{matrix}e=0 \\ 2s=0 \\ nr=r^n \\ sr+r=sr^2=r+sr \\ sr+sr=r^2 \\ 6r=r^6=0=e \end{matrix}$$

    This basically uses the addition from $C_2\times C_6$
    Last edited by SlipEternal; Aug 14th 2018 at 05:00 PM.
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    Re: Two binary operations on a group

    Quote Originally Posted by SlipEternal View Post
    How about

    $$\begin{matrix}e=0 \\ 2s=0 \\ nr=r^n \\ sr+r=sr^2=r+sr \\ sr+sr=r^2 \\ 6r=r^6=0=e \end{matrix}$$

    This basically uses the addition from $C_2\times C_6$
    If I understand what you posted you are giving the same group both addition and multiplication that works out to be the same? Or did I miss something. Yeah, I guess that works though I was hoping for more of a construction method taking the definitions from a multiplicative group and working out an addition table such that we get a ring. ie. The distributive property. That's what you've done but I was hoping for something less "trivial." Perhaps no method exists... I'm not any good with the theory yet. It may not be possible to do in general.

    Thanks!

    -Dan
    Last edited by topsquark; Aug 14th 2018 at 05:18 PM.
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    Re: Two binary operations on a group

    Quote Originally Posted by topsquark View Post
    If I understand what you posted you are giving the same group both addition and multiplication that works out to be the same? Or did I miss something. Yeah, I guess that works though I was hoping for more of a construction method taking the definitions from a multiplicative group and working out an addition table such that we get a ring. ie. The distributive property. That's what you've done but I was hoping for something less "trivial." Perhaps no method exists... I'm not any good with the theory yet. It may not be possible to do in general.
    A ring $\mathcal{R}$ is a set with two operations $\large{+~\&~\cdot}$
    There are eight axioms that must be followed:
    $ \begin{align*}1&: (\{a,b\}\subset\mathcal{R})[a+ b\in\mathcal{R}]\\2&: (\{a,b\}\subset\mathcal{R})[a+ b=b+a]\\ 3&: (\{a,b,c\}\subset\mathcal{R})[(a+ b)+c=a+(b+c)]\\ 4&: (\exists 0\in\mathcal{R})(\forall a \in\mathcal{R})[a+0=a]\\ 5&: ( \forall a \in\mathcal{R})(\exists-a\in\mathcal{R})[a+(-a)=0]\\ 6&: (\{a,b\}\subset\mathcal{R})[a\cdot b\in \mathcal{R}]\\7&: (\{a,b,c\}\subset\mathcal{R})[(a\cdot b) \cdot c=a\cdot (b\cdot c)] \\8&: (\{a,b,c\}\subset\mathcal{R})[a\cdot( b+c)=a\cdot b+a\cdot c]\end{align*}$

    Even though it is tedious in the extreme to prove these five theorems it is very instructive.
    $ \begin{align*}\text{Theorem }1&: (\forall a\in\mathcal{R})[a\cdot 0=0\cdot a=0] \\\text{Theorem }2 &: (\forall a\in\mathcal{R})[-a\text{ is unique}]\\\text{Theorem }3&: (\forall a\in\mathcal{R})[-(-a)=a]\\\text{Theorem }4&: (\forall a\in\mathcal{R})(\forall b\in\mathcal{R})[(-a)\cdot b=a\cdot(-b)=-(a\cdot b)]\\\text{Theorem }5&: (\forall a\in\mathcal{R})(\forall b\in\mathcal{R})[(-a)\cdot(-b)=a\cdot ]\end{align*}$
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