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redundancy of two-sidedness in definition of group

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\begin{document}
In the definition of group, one usually supposes that there is a two-sided identity element and that 
any element has a two-sided inverse (cf. \PMlinkname{group}{Group}).

The group may also be defined without the two-sidednesses:\\

A {\it group} is a pair of a non-empty set $G$ and its associative binary operation 
$(x,y) \mapsto xy$ such that

1) the operation has a right identity element $e$;

2) any element $x$ of $G$ has a right inverse $x^{-1}$.\\

We have to show that the right identity $e$ is also a left identity 
and that any right inverse is also a left inverse.

Let the above assumptions on $G$ be true.\, If $a^{-1}$ is the right 
inverse of an arbitrary element $a$ of $G$, the calculation
$$a^{-1}a = a^{-1}ae = a^{-1}aa^{-1}(a^{-1})^{-1} = 
a^{-1}e(a^{-1})^{-1} = a^{-1}(a^{-1})^{-1} = e$$
shows that it is also the left inverse of $a$.\, Using this result, 
we then can write
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a,$$
whence $e$ is a left identity element, too.

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