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I have a homework question and I don't know how to approach this exercise.

The exercise is the following:

Let's suppose $G$ is a set with binary function * defined for its members, which is:

  • closings;
  • associative;
  • there's $e\in G$, so that $a\star e=a$ where $a\in G$;
  • for each $a\in G$, there's a $b\in G$ so that $a\star b=e$.

Prove that $G$ is a group.

I have no Idea how approach this exercise.

pay attention that 3,4 are Noncommutative.

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    $\begingroup$ What is the definition of a group? $\endgroup$ – user41442 Apr 18 '13 at 10:19
  • $\begingroup$ @user41442 There's no definition $\endgroup$ – Billie Apr 18 '13 at 10:22
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    $\begingroup$ Every mathematical structure has a definition, so you better find that definition. Try wikipedia for example. Then you can use that definition to see why your set G is a group. $\endgroup$ – Applied mathematician Apr 18 '13 at 10:26
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    $\begingroup$ @user1798362: this could well be the definition of a group, but assuming there is something to prove, the definition you got in class should be that 3,4 are two sided, i.e there exists $e\in G$ such that $a*e=e*a=a$ for all $a\in G$ and for any $a \in G$ there exists $b\in G$ such that $a*b=b*a=e$ $\endgroup$ – Dennis Gulko Apr 18 '13 at 10:42
  • $\begingroup$ This seems to be the same question as Right identity and Right inverse implies a group. $\endgroup$ – Martin Sleziak Oct 4 '13 at 14:05
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First, let $a\in G$. By (4), there exists $b\in G$ such that $a*b=e$. Moreover, again by (4), there exists $c\in G$ such that $b*c=e$. Then: $$a=a*e=a*(b*c)=(a*b)*c=e*c$$ And thus $$b*a=b*(e*c)=(b*e)*c=b*c=e$$ Which proves the second direction of (4).
This proof is independent of the finiteness of $G$.
Now you should be able to complete the proof (you still need to show that $e*a=a$)

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This result is true if $G$ is a finite set, indeed:

Fix $a\in G$ and define $\varphi_a:G\to G$, $x\mapsto x*a$ then if $\varphi_a(x)=x*a=\varphi_a(y)=y*a$ then multiply on RHS by $b$ and we find $x=y$ hence $\varphi_a$ is injective and by finite cardinality of $G$, $\varphi_a$ is bijective.

Now there's $b'\in G$ such that $\varphi_a(b')=b'*a=e$ and multiply on the RHS by $b$ we find $b'=b$ so $b*a=a*b=e$

Finaly there's $e'$ such that $\varphi_a(e')=e'*a=a$ then we multiply on the RHS by $b$ and we have $e'=e$ so $e*a=a*e=a$.

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By assumption, $e$ is a right-neutral element for $*$ and, given $a\in G$, the element $a'\in G$ such that $a*a'=e$ is a right-inverse for $a$.

All what you have to prove is that $e$ is also a left-neutral element and that $a'$ is also a left inverse of $a$.

From $a*a'=e$ you can deduce: $$e*(a*a')=e*e=e$$ and by associativity $$(e*a)*a'=e=a*a'$$

Let $a''$ be the right-inverse of $a'$. Then

$$[(e*a)*a']*a''=(a*a')*a''$$ Thus $$(e*a)*(a'*a'')=a*(a'*a'')$$ $$(e*a)*e=a*e$$ $$e*a=a$$ Again from $a*a'=e$ you deduce $$a'*(a*a')=a'*e$$ $$(a'*a)*a'=a'$$ $$[(a'*a)*a']*a''=a'*a''$$ $$(a'*a)*(a'*a'')=e$$ $$(a'*a)*e=e$$ $$a'*a=e$$

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