This is part of Exercise 2.6.6 of Howie's "Fundamentals of Semigroup Theory". I apologise in advance if this is a duplicate.
The Details:
Let $S$ be a semigroup.
Definition 1: We call $S$ right simple if $\mathcal R=S\times S$, where $\mathcal R$ is Green's $R$-relation.
Definition 2: We call $S$ left cancellative if $(\forall a, b, c\in S) ca=cb\implies a=b$.
Definition 3: If $S$ is right simple and left cancellative, then $S$ is by definition a right group.
Definition 4: If for all $a, z\in S$, $az=z$, then $S$ is a right zero semigroup.
The Question:
Show that if $G$ is a group and $E$ is a right zero semigroup, then the direct product $G\times E$ is a right group.
My Attempt:
Let $(g, e), (h, f)\in G\times E$. Then $$\begin{align} (g, e)G\times E&=\left\{(gi, d)\mid (i, d)\in G\times E\right\} \\ &=G\times E \end{align}$$ and similarly $(h, f)G\times E=G\times E$, so $(g, e)\mathcal R (h, f)$. Thus $\mathcal R=S\times S$, so $S$ is right simple.
Suppose $(g, d), (h, e), (i, f)\in G \times E$ such that $$ (i, f)(g, d)=(i, f)(h, e). $$ Then $ig=ih\implies g=h$ and $d=fd=fe=e$, so $S$ is left cancellative.
Hence $G\times E$ is a right group.
Correct. $\def\x{ }$