# Definition of 'Abelian group'

#### QuestForInsight

##### Member
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
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

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#### QuestForInsight

##### Member
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
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?

#### QuestForInsight

##### Member
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.

#### QuestForInsight

##### Member
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
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

#### QuestForInsight

##### Member
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
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
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
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...