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Homeredundancy of two-sidedness in definition of group

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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. group).

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

A 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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