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Let $G$ be a semigroup such that $\forall a\in G$, $\exists b \in G$ such that $a = aba$ and $\exists! e \in G$ such that $e^2=e$. Prove that $G$ is a group.

This is an exercise problem from my introductory Abstract Algebra course and we've just started groups and subgroups.

Since we already know $G$ is a semigroup, I only need to show that $G$ has an identity and every element of $G$ has an inverse. However, I'm not really sure on how to do that.

We also had a theorem which says that

If $\forall a,b \in G$, $\exists x,y \in G$ such that $ax=b$ and $ya=b$, then $G$ is a group.

But I don't think I can't use it here.

Any hints would be much appreciated. Thanks

Zev Chonoles
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Sai
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1 Answers1

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Given $a\in G$, take $b\in G$ such that $a=aba$. So $$ab=(ab)(ab)=(ab)^{2}$$ and $$ba=(ba)(ba)=(ba)^{2},$$hence $ab=ba=e$.

Now, note that $$ae=a(ba)=aba=a$$ and $$ea=(ab)a=aba=a.$$ Thus, we prove that $e$ is the identity element and every element has an inverse, i.e., $G$ is a group.

Rafael
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