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#### QuestForInsight

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- Jul 22, 2012

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Is the above definition correct/does it miss anything?

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- Jul 22, 2012

- 35

Is the above definition correct/does it miss anything?

- Jan 26, 2012

- 890

Presumably you mean: Let \( (G,+) \) be a group, or provide some other definition for "\(+\)" (like identifying the binary "\(+\)" operation with \(f( .,. )\)) which appears unannounced above.

Is the above definition correct/does it miss anything?

Also you appear to have required the existence of a universal inverse (in that the existence of \( \xi' \) is prior to that of \( \xi \)).

This is supposed to define a binary operation "\(+\)" on \(\mathbb{G}\) using the the function \(f(.,.)\), which is closed. Then require it to be commutative, and that there is a element \( \omega \) of \(\mathbb{G}\) that behaves like the identity under "\(+\)", and that each element has an inverse under "\(+\)".

My prefernce would be for something like:

Let \( (G,\oplus) \) be a group such that \(\forall \xi, \eta \in \mathbb{G}:\ \xi \oplus \eta=\eta \oplus \xi\). This structure is called an Abelian group.

CB

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- Jul 22, 2012

- 35

Could you elaborate on the above bit, please?Also you appear to have required the existence of a universal inverse (in that the existence of \( \xi' \) is prior to that of \( \xi \)).

Cool. Can we tweak my attempt to give a definition of 'group' for that matter?Let \( (G,\oplus) \) be a group such that \(\forall \xi, \eta \in \mathbb{G}:\ \xi \oplus \eta=\eta \oplus \xi\). This structure is called an Abelian group.

CB

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- Feb 7, 2012

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Is there anything in that definition to ensure that the operation given by $\xi+\eta = f(\xi, \eta)$ is associative?

Is the above definition correct/does it miss anything?

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- Jul 22, 2012

- 35

I was thinking that commutativity implies associativity? Now that I think about it, though, I'm not sure.Is there anything in that definition to ensure that the operation given by $\xi+\eta = f(\xi, \eta)$ is associative?

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- Jul 22, 2012

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Okay now?

- Jan 26, 2012

- 890

Why are you trying to do this? What is wrong with the usual group axioms?Revised Definition:Let $\mathbb{G}$ be a set with a map $(\xi, ~ \eta) \mapsto f(\xi, ~\eta)$ from $\mathbb{G}\times\mathbb{G}$ into $\mathbb{G}$. For every pair $(\xi, ~ \eta)$ in $\mathbb{G}$ let $f(\xi, ~\eta) = f(\eta, ~ \xi)$; every triplet $(\xi, ~ \eta, ~\zeta)$, let $f((\xi, ~ \eta), ~ \zeta) = f(\xi, ~ (\eta, ~ \zeta))$ -- and so on. Suppose there are elements $\omega$ and $\xi'$ in $\mathbb{G}$ such that for every $\xi$ in $\mathbb{G}$ we have $f(\xi, ~ \omega) = f(\xi)$ and $f(\xi, ~ \xi') = f(\omega).$ This structure is called an Abelian group when $f(\xi, \eta) = \xi+\eta$ (assuming we defined addition).

Okay now?

Also: $f((\xi, ~ \eta), ~ \zeta) = f(\xi, ~ (\eta, ~ \zeta))$ is meaningless, may be you want:

$f(f(\xi, ~ \eta), ~ \zeta) = f(\xi, ~ f(\eta, ~ \zeta))$

CB

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- Jul 22, 2012

- 35

Nothing wrong with the usual group axioms. I was just trying this out.Why are you trying to do this? What is wrong with the usual group axioms?

Also: $f((\xi, ~ \eta), ~ \zeta) = f(\xi, ~ (\eta, ~ \zeta))$ is meaningless, may be you want:

$f(f(\xi, ~ \eta), ~ \zeta) = f(\xi, ~ f(\eta, ~ \zeta))$

CB

Why is that notation meaningless, though?

Say we have f(a, b) = a+b. If b = (c, d), can't we write a+(c+d) as f(a, (c, d))?

- Feb 29, 2012

- 342

- Jan 26, 2012

- 890

\(f\) is a function on \( \mathbb{G}\times \mathbb{G} \), \((a,b) \) could be anything, but you probably intented it to represent an ordered pair, that is an element of \( \mathbb{G}\times \mathbb{G} \), so ((a,b),c) is in \( (\mathbb{G}\times \mathbb{G}) \times \mathbb{G}\) which is not in the domain of \(f\).Nothing wrong with the usual group axioms. I was just trying this out.

Why is that notation meaningless, though?

Say we have f(a, b) = a+b. If b = (c, d), can't we write a+(c+d) as f(a, (c, d))?

What you need for associativity is on the last line of my previous post.

CB

Using his notation allows you to separate the set from the group. I mean, a binary operation *is* a function taking an ordered pair to a single element. Thinking about it this way allows you to define other structures on a well-known set with perhaps less confusion (and encourages students to not think of the groups as just the given set, perhaps?). For example, the Tropical semi-field is the Real numbers (with or without infinity) where addition is $\max(a, b)$ and multiplication is $a+b$.Why are you trying to do this? What is wrong with the usual group axioms?

Getting addition and multiplication mixed up seems to be a common theme of any tropical talk...