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(Reference : Fraleigh, A first course in abstract algebra)

Prove that a nonempty set $G$, together with an associative binary operation * on $G$ such that $a*x=b$ and $y*a=b$ have solutions in G, $\forall a,b\in G$, is a group.

Because * is an associatie operation, we have to prove that: there exists $e\in G$ such that $e*a=a=a*e$ for any $a\in G$, and that for any $a\in G$ there exists $\alpha \in G$ such that $a*\alpha=e=\alpha *a$.

This is very confusing, first I tried to prove the identity exists. If we consider a=b, then $a*x=a=y*a$, and because of the hypothesis $x,y\in G$ always exists, now we have to show that $x=y$, but I just can't do this with out inverses, so I tried to prove that the inverses exists, however I have no idea how the identity looks like, I'm stuck.

And I tried to get unstuck, so I thought particular cases, first if $G$ had only one element, then what would happend if it had just two elements... but I remembered that we don't know if G is finite or not, and I think this makes things a little bit more complicated.

What can I do?

HK Lee
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Abodi
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    Not only do you need to show that $x=y$, but you would need to show that the same $x$ works for all elements of $G$, not just the particular element $a$. – Braindead Jan 30 '14 at 06:02
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    If you require uniqueness of the solutions, see this question: http://math.stackexchange.com/questions/346595/prove-that-g-circ-is-a-group-if-a-circ-x-b-and-x-circ-a-b-have-uni Some discussion whether uniqueness is needed can be found here: http://math.stackexchange.com/questions/488783/uniqueness-of-solutions-of-ax-b-and-ya-b-in-a-semigroup and also in this answer: http://math.stackexchange.com/questions/346595/prove-that-g-circ-is-a-group-if-a-circ-x-b-and-x-circ-a-b-have-uni/346661#346661 – Martin Sleziak Jan 30 '14 at 07:02
  • Thanks @MartinSleziak those references where very helpful... – Abodi Jan 31 '14 at 05:04

2 Answers2

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Let $e(a)$ is the solution of $ax=a$. Then for any $b$ there exists $y$ such that $ya=b$. Then $be(a)=yae(a)=ya=b$. So $e=e(a)$ is a right unit. And the equation $ax=e$ is always solvable, so $G$ has right unit and right inverse. Then it is a group.

user68061
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$\forall a$, there exists $x$ and $y$ such that $ax = a$ and $ya = a$.

$\Rightarrow$ $xx = x$ and $yy = y$ (because $\exists$ ($t_x$ and $t_y$) such that $t_x \cdot a = x$ and $a \cdot t_y = y$).

$\Rightarrow$ $yx = y$ and $yx = x$ (because $\exists$ ($m$ and $n$) such that $m \cdot x = y$ and $y \cdot n = x$).

$\Rightarrow$ $x = y$.

($xx = x$) $\Rightarrow$ $\forall a$, $ax = x$ and $xa = a$ (because $\forall a$ $\exists$ ($c$ and $d$) such that $cx = a$ and $x \cdot d = a$).

Also, $\forall a$, $\exists c$ such that $ac = x$.

This implies $cac = cx = c$ (since $ax = a$ for all $a$).

Hence, $cax = x$ (since there exists a $d$ such that $cd = x$).

Therefore, $ca = x$.

Thus, there exists an identity element $x$ which is both right and left identity and an inverse element $c$ which is right as well as left inverse.

vib_29
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