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The definition of a finite group is : A group $(G,+)$ is a finite group under arbitrary group operation "+" (not an addition) if "$G$" is a finite set. But my question is, if $(G,+)$ is a finite group then necessarily "$G$" is a finite set? If yes then how can we prove it and if the answer is no then what is the counterexample of it?

Shaun
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Keshav shrestha
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    this is a matter of definition. A "finite group" is a finite set equipped with a group structure. – lulu Jan 18 '23 at 12:33
  • @lulu so obviously a finite group (G,+) implies the group G is a finite set right? – Keshav shrestha Jan 18 '23 at 12:39
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    As I say (and as you said in the post), that's just part of the definition. It's like saying that describing a figure as a "right triangle" implies that it is a triangle. – lulu Jan 18 '23 at 12:43
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    @lulu Taking the statement of that definition at face value, using only the logic that's explicitly present in the statement, this is an entirely legitimate question. – Arthur Jan 18 '23 at 12:46
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    @Arthur Whether such questions are "legitimate" is a matter of taste. What always confuses me in such cases is that such questions regularly get answers after everything relevant has been said in the comments. I do not get the point of this practice. – Peter Jan 18 '23 at 12:54
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    Assuming my answer does get at the actual core of OP's issue, how has "everything relevant" been said in the comments? Also, I do not get the point of the practice of answering question in the comment section. – Arthur Jan 18 '23 at 12:59
  • Not sure why this has become controversial. To say that something follows at once from the definition is not to say that it is illegitimate to ask about it. Words and terms are sometimes used suggestively rather than literally. A "linear polynomial" is usually not a "linear function", confusingly. In this case, the word "finite" was used literally but there's no harm in asking about it. – lulu Jan 18 '23 at 13:02
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    @lulu My point is, if you don't know about the convention I talk about in my answer, it doesn't even follow from the given definition that a finite group has a finite underlying set. And that makes it all the more suitable to actually ask about. – Arthur Jan 18 '23 at 13:04
  • @Arthur I would say, rather, that it is crucial to make it clear when something is a definition and not a Theorem or merely an example. Writing "Alice is a mathematician" defines neither Alice nor mathematicians. I agree that people often write unclearly...writing "$f(x)=ax+b$ is a linear polynomial" without clearly indicating whether you are defining a linear polynomial that way or simply giving an example. That can certainly be confusing. – lulu Jan 18 '23 at 13:12
  • @lulu I challenge you to go find a textbook right now, pick a definition at random, and see whether 1) it uses "if" or "if and only if", and 2) It ought to use "if" or "if and only if". I think you will find most of the time that it uses "if", but ought to use "if and only if". For instance, "A group $(G, \cdot)$ is abelian if $a\cdot b = b\cdot a$ for all $a, b\in G$" doesn't really say that all abelian groups commute. Or the example in the OP. – Arthur Jan 18 '23 at 13:16
  • @Arthur Sure. I'd say all that was subsumed by declaring that what follows is a Definition (and not an Example or a Theorem or whatever). But I agree that there's no harm in clarifying the point. – lulu Jan 18 '23 at 13:25

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Definitions are really if-and-only-if statements, but they are phrased using just if, because that's how the convention turned out. So really, it should have been

A group $(G,+)$ is a finite group if and only if $G$ is a finite set.

but that's just not how definitions are phrased in modern mathematical parlance. We instead say

A group $(G,+)$ is a finite group if $G$ is a finite set.

and expect readers to understand that because this is a definition, they need to read it differently from how they would read that exact same phrase if it had been a theorem.

Arthur
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  • so each and every almost all definition are really "if and only if "statements and following convention or abbreviately the definitions are phrased using just "if" but following those convention these definitions does not lead to false understanding about the definitions? – Keshav shrestha Jan 18 '23 at 12:59
  • @Keshavshrestha Yes, more or less. It is just implicitly understood that not only are groups with finite sets called finite groups, but also when you say you have a finite group, you mean that the underlying set is finite. This is so implicitly understood and ingrained that several people posting under your question post didn't even catch that that could potentially be an issue. – Arthur Jan 18 '23 at 13:01
  • Regarding to the definition of open set in real analysis A set "S" is an open set if and only if there exists an open ball B(x,r) such that B(x,r) is entirely contained in a set S. But what we found the definition of open set in convention manner is : A set "S" is an open set if there exists an open ball B(x,r) such that B(x,r) is entirely contained in a set "S". so both definitions are equally valid? – Keshav shrestha Jan 18 '23 at 13:08
  • @Keshavshrestha Those two sentences are entirely identical apart from the " arount the final S. Maybe you meant one of them to be A set "S" is an open set if there exists an open ball B(x,r) such that B(x,r) is entirely contained in a set "S"? And yes, apart from the fact that it's not the exact definition of an open set (you need something more about the $x$ in there, for instance), as a definition, it is entirely according to current style to write it the way I wrote it here. – Arthur Jan 18 '23 at 13:11
  • yes i mean "S" .So those both definitions are equally valid and identical right? But to prove some theorem we should clarify between "if" statements and "if and only if" statements according to theorem yes? – Keshav shrestha Jan 18 '23 at 13:15
  • @Keshavshrestha Yes. But if you had had a different definition of open sets, and then a theorem that said "A set $S$ is an open set if there exists an open ball $B(x,r)$ such that $B(x,r)$ is entirely contained in a set $S$", then that theorem doesn't say that open sets necessarily contain the open balls that we all know they do. So the theorem becomes weaker because it doesn't have the "if and only if", even though a definition wouldn't need it. (Again, that definition of open sets is incomplete, I'm just continuing to use it because correcting it is not in scope of this discussion.) – Arthur Jan 18 '23 at 13:17
  • thank you for your lucid answer so finally a simple misleading of the word "if" and "if and only if" in theorem or in definitions could lead us to disastrous,false and wrong understanding right? – Keshav shrestha Jan 18 '23 at 13:25
  • @Keshavshrestha I might not go so far as calling it "disastrous", but yes, while theorems mean exactly what they say and not anything more, definitions mean a little more than they say. And this is something one just has to know. – Arthur Jan 18 '23 at 13:29
  • yes just like you give an example how misleading could be a definition be if we use definition for granted as a group (G,⋅) is abelian if a⋅b=b⋅a for all a,b∈G instead of using the definition as a group(G,.) is an abelian if and only if a.b=b.a for all a,b belongs to G. – Keshav shrestha Jan 18 '23 at 13:45
  • If and only if $a\iff b$. usually requires that both $(a \to b)$, and $(b \to a)$ are true. Definitions need to satisfy this as well. – amWhy Jan 18 '23 at 18:38
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    @amWhy They do. But they usually aren't written that way. Because convention. – Arthur Jan 18 '23 at 21:23
  • There is no such convention, @Arthur. Besides, please refrain from answering questions void of sufficient context to successfully answer. – amWhy Jan 18 '23 at 22:27
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    @amWhy Isn't there such a convention? Atiyah-MacDonald: "An ideal $p$ in $A$ is prime if $p\neq (1)$ and if $xy\in p\implies x\in p$ or $y\in p$". Hartshorne: "A subset $Y$ of $\Bbb A^n$ is an algebraic set if there exists a subset $T\subseteq A$ such that $Y=Z(T)$". I dare you to find examples that explicitly use the proper "if and only if". And I fail to see how this question lacks substance. Misunderstanding this convention is in my opinion a substantive problem of the kind this site is made for helping with. – Arthur Jan 19 '23 at 06:05
  • But @Arthur i have a little doubt here, if "G is a finite set then it necessarily doesnot imply (G,.) a finite group because set "G" has to satisfy the axiom of the group but if (G,.) is a finite group then it implies that the underlying set "G" is a finite set so we can't use if and only if for this definition, isn't that right please correct me if i am wrong. – Keshav shrestha Jan 21 '23 at 14:56