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Some books say stuff like "if $\forall x \in G$, $x^2=e$, then $G$ is abelian". But the notation of a group is $\langle G, \circ \rangle$, and that looks like an ordered pair. So, should not the elements of the group be $\{G\}$ and $\{\circ,G\}$ by definition of ordered pair? Or am i getting wrong the notation?

I got this doubt also with the notation of partially ordered sets.

Schonfinkel
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Gustavo Alejandro Castellanos
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    It's a language abuse, the book should say that "if $\forall x \in G$, $x^2=e$, then $\langle G, \circ\rangle$ is abelian". – Git Gud Feb 05 '17 at 19:13
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    It's been my experience that most books will introduce a group as $\langle G, \circ \rangle$ and then say something to the effect of "if the operation is clear from context then we'll refer to the group simply as $G$..." –  Feb 05 '17 at 19:55
  • I thought ordered pairs are sets in set theory? For example, the pair $(a, b)$ could be the set ${{a}, {a, b}}$...? – user541686 Feb 05 '17 at 21:55
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    @GitGud: I agree that it is a language abuse, but why would the book use such obtuse notation? In my opinion, it should say "If $x^2=e$ for all $x$ in $G$, then $G$ is abelian." There is nothing wrong with abuse of language, as long as it does not cause too much confusion (compared to how convenient it might be). And identifying a structure with its underlying set in notation is probably one of the most universal abuses of language in mathematics. Writing quantifiers instead of words (in proper mathematical writing) is a much worse offence (in my opinion). – tomasz Feb 06 '17 at 03:01
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    Also, as a side comment: an ordered pair is simply an object which consists of two parts: the first and the second element. In set theory, it can be modelled according to one of the many possible constructions, but you should not confuse these constructions with the definition of an ordered pair, which is more or less what I wrote in the first sentence. The only real point of the construction is to show that it can all be modelled (or interpreted) completely in language of set theory, which can be useful in some arguments (or in building foundations), but not in basic group theory. – tomasz Feb 06 '17 at 03:08
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    In other words, be careful when someone tells you that $3\in 4$ (though this is an useful abuse of notation in some contexts, so don't discard it outright), don't believe anyone saying that ${3}\in (3,4)$, and stay away from people telling you that $\frac{3}{1}\in \pi$. – tomasz Feb 06 '17 at 03:11
  • @tomasz You misunderstood me, or maybe I wasn't clear enough. What I meant was something like "if the author wanted to be 100% correct, he should have written ...". I'm not ready to commit to the view that $(G,\circ)$ should be used on all instances. – Git Gud Feb 06 '17 at 07:24
  • @tomasz Regarding the ordered pair definition, I believe we have different definitions for the word definition $\ddot \smile$ "First and second element" isn't a definition in my book, it's a characteristic of the concept of ordered pair. This concept "in set theory, can be modelled according to one of the many possible constructions, but you should not confuse these constructions with the" concept of an ordered pair. – Git Gud Feb 06 '17 at 07:26
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    And I will believe that ${3}\in (3,4)$ because I'm one of the people that claims it. The fact that there are many ways to model the concept of ordered pair in $\sf ZFC$ doesn't change the fact that there is one way which is much more popular than the others (for purely social reasons). It's not like the number $e$ which really has a lot of equally popular definitions. – Git Gud Feb 06 '17 at 07:27
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    @GitGud: I think a significant majority of mathematicians would agree that saying that ${3}\in(3,4)$ is nonsense. But here we'll just have to agree to disagree. – tomasz Feb 06 '17 at 12:19

3 Answers3

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Yes, a group is an ordered pair: the first element of the pair is a set (the underlying function of the group), and the second is a binary function on that set (which, in set theory, is actually a set too).

Saying something like "$G$ is abelian" is an abuse of notation: technically it's incorrect, but it has only one reasonable interpretation (this is only true btw if we aren't considering two different group structures on the same set, which we sometimes do). It's used because it's slightly easier to write than "$(G, \circ)$ is abelian."

Incidentally, given that "$e$" isn't actually part of the tuple $(G, \circ)$, that's also an abuse of notation - one should write $e_G$ (to distinguish it from the identity of some other group) or similar. But, again, we can get away with it in contexts where it won't lead to confusion. Also, it's worth pointing out that many texts treat groups as ordered triples of the form $(G, \circ, e)$.

Noah Schweber
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    I disagree with the third paragraph partially. Many times $e$ isn't explicitly present in the definition of group and rather it is required there exists an element in $G$ that behaves a certain way. – Git Gud Feb 05 '17 at 19:37
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    @GitGud To use the symbol "$e$" without it having been defined already is, indeed, an abuse of notation. $G$ being a group of course implies that it has an identity; however, that doesn't tell us what that identity is called. Technically, we have no way of knowing what "$e$" refers to. And if we assume "$e$" is the standard notation for the identity of a group - which is true - then what do we do if we are dealing with two different groups at once? Again, this is a perfectly benign abuse of notation, but it is technically an abuse of notation. – Noah Schweber Feb 05 '17 at 19:44
  • @GitGud "Many times $e$ isn't explicitly present in the definition of group" I never said otherwise; all I said is that many times it is. Incidentally, for an example of where including $e$ in the explicit definition of a group is really important, btw, look at universal algebra - in the language without $e$, groups aren't a variety! – Noah Schweber Feb 05 '17 at 19:44
  • I agree, and that's why I said disagreed partially. Certainly, the way the OP formulated the question, $(G,\circ , e)$ should be used. – Git Gud Feb 05 '17 at 20:07
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    @GitGud This reminds me of possibly one of my favorite facts. If you have an associative monoid $(G,)$ with right and left division and cancelation (i.e. $\forall;a,b\exists !c,d (ac=b)\wedge (da=b)$ then there exists an identity with respect to $$ and $G$ is a group.:) I just couldn't help myself because it seems incredible that a local property implies a global unit. – DRF Feb 05 '17 at 21:04
  • @DRF: And you don't even need to assume uniqueness of $c$ and $d$. – hmakholm left over Monica Feb 06 '17 at 00:17
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Often, in maths you encounter structures that are sets plus some structure on that set. For example with groups, you are dealing with a set $G$ and some operation $\cdot: G \times G \to G$ on this set. In topology, you have a set $X$ and a topology $\tau$ on this set. In measure theory you have a set $X$, a $\sigma$-algebra on $X$, denoted $\Gamma$ and a measure on $\Gamma$ denoted $\mu$.

In all of these cases you can describe the thing properly by providing the set and the structure together: a group is $(G,\cdot)$, a topological space is $(X, \tau)$ and a measure space is $(X, \Gamma, \mu)$. I think this is what your notation is about.

Matias Heikkilä
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This is technically an abuse of language. Conflating a tuple $(A,\ldots)$ defining a set with structure with the set $A$ itself extremely common and I don't remember hearing anyone ever complain about it. This is because the ordered tuple construction is quite artificial and is not the only way to associate a set with the structures we put on it.

For the example of groups, we could instead define a group as an object in the category $\mathbf{Grp}$ of all groups. Now this really is a set. The algebraic structure is then hidden in the morphisms of the category.

Matt Samuel
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  • I complained about it when I first learned about algebraic structures. The terminology "a set $A$ together with an operation $*$" was very unclear to me because I couldn't tell that "together" was short for considering it as an ordered pair, I simply didn't have enough mathematical maturity to see this. This prompted me to to deal with issues such as $\mathbb Z$ both being and not being a group, after all multiplication and addition is "together with the integers", in some sense. – Git Gud Feb 05 '17 at 19:33
  • @GitGud Well I guess I must not have been there to hear your complaint then! – Matt Samuel Feb 05 '17 at 19:34
  • $\ddot \smile{}$ – Git Gud Feb 05 '17 at 19:35