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Definition of 'Abelian group'

Jul 22, 2012
35
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 $f(\xi, ~\eta) = f(\eta, ~ \xi)$. 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 $f(\xi, \eta) = \xi+\eta$.

Is the above definition correct/does it miss anything?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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 $f(\xi, ~\eta) = f(\eta, ~ \xi)$. 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 $f(\xi, \eta) = \xi+\eta$.

Is the above definition correct/does it miss anything?
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.

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
 
Last edited:
Jul 22, 2012
35
Thank you.

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

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
Cool. Can we tweak my attempt to give a definition of 'group' for that matter?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
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 $f(\xi, ~\eta) = f(\eta, ~ \xi)$. 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 $f(\xi, \eta) = \xi+\eta$.

Is the above definition correct/does it miss anything?
Is there anything in that definition to ensure that the operation given by $\xi+\eta = f(\xi, \eta)$ is associative?
 
Jul 22, 2012
35
Is there anything in that definition to ensure that the operation given by $\xi+\eta = f(\xi, \eta)$ is associative?
I was thinking that commutativity implies associativity? Now that I think about it, though, I'm not sure.
 
Jul 22, 2012
35
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?
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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?
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
 
Jul 22, 2012
35
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
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))?
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
I believe what Cap means is that if you design $f(c,d) = c+d$, then $(c,d)$ is just the pair of elements you're applying the operation, not the result of the operation, an element of the space itself. By writing $f(a,(c,d)) = a+(c+d)$ you're instead accepting a pair instead of a single element, I'd go as far as to say it's an abuse of notation. The correct is $f(a,f(c,d)) = a+(c+d)$, like pointed out.
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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))?
\(f\) 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 \(f\).

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

CB
 

Swlabr

New member
Feb 21, 2012
27
Why are you trying to do this? What is wrong with the usual group axioms?
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 $a+b$.

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