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Let $G$ be a nonempty set closed under an associative product, which in addition satisfies:

(a) There exists an $e\in G$ such that $a· e = a$ for all $a\in G$.

(b) Give $a\in G$, there exists an element $y(a)\in G$ such that $a·y(a) = e$. Prove that $G$ must be a group under this product.

My solution goes like this:

First we need to verify the closure property i.e if $a,b\in G $ then $a.b\in G$, which is true by definition. Next, we need to verify the associative property i.e if $a,b,c\in G$ then , $a.(b.c)=(a.b).c$. Next , we verify the inverse property , i.e for every $a\in G$ ,$\exists y(a)\in G $ such that $a.y(a)=e$ ,which means $y(a).a.y(a)=y(a).e=y(a)$. Now, $y(a)\in G $, hence $\exists y(y(a))\in G $ such that $y(a).y(y(a))=e$. Thus, $y(a).a.y(a).y(y(a))=y(a).y(y(a))=e$, which means , $(y(a).a).y(a).y(y(a))=y(a).y(y(a))=e$, hence $y(a).a=e$. Thus $$y(a)\in G $$ is the inverse for every $a\in G $ . Next, we verify the identity property . Given, $\forall a \in G $,$\exists e\in G$ such that $a.e=a$ which means $a.y(a).a=a$, hence $(a.y(a)).a=a$ which means $e.a=a$, hence $e\in G $ is the identity element. Hence, $G $ is a group under this binary operation.

Is the above proof valid? I know that there are numerous threads relating this post. But ,is the above proof correct? If not, then why? I just want to verify my proof...

Shaun
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Arthur
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1 Answers1

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  1. I suggest removing the part with closure property entirely.

  2. You're saying that we need to verify the associative property, but I'd just say that it's associative by assumption.

  3. After saying that we need to verify the inverse property you say "i.e." which is Latin for id est, translating to English, that is. This means you're saying that $$\forall_{a\in G}\exists_{y(a)\in G} a\cdot y(a) = e$$ is the inverse property, which is not true, because we need to demand $y(a)\cdot a = e$ as well. Moreover, we want to fix and element $e\in G$ for which $\forall_{a\in G} a\cdot e = a$ beforehand, which you didn't do. Even after fixing all of the things above, you'd like to avoid calling it the inverse property, since you haven't proven that $e$ is the identity element just yet.

  4. I suggest writing $a' = y(a)$ and $a'' = y(a')$ to save up on writing.

  5. I'd write $a'a = a' a a' a'' = a'a'' = e$ to make it clearer what you're using.

Jakobian
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  • Yes! I do agree with u that the definition of inverse property...I didn't completely write , I should add $y(a).a=e$. Also, ur saying I must add that $e$ is assumed to be the identity element of $G$, right?Actually , in that statement "Given $\forall a \in G$ ,$\exists e \in G$ such that $a.e=a$..." , I was writing what was given in the question and not writing about the inverse property...and yes we should write $e$ as $e'$ in the inverse property case and then when we prove $e $ to be the inverse then we can say $e=e'$, right? As , we see that $e$ exists as an identity element... – Arthur Oct 13 '22 at 13:55
  • Your proof is okay, just we don't know that the property is the inverse property just yet. After we prove that $e$ is the identity we are free to call it that of course – Jakobian Oct 13 '22 at 14:38