# redundancy of two-sidedness in definition of group

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 (http://planetmath.org/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

$${a}^{-1}a={a}^{-1}ae={a}^{-1}a{a}^{-1}{({a}^{-1})}^{-1}={a}^{-1}e{({a}^{-1})}^{-1}={a}^{-1}{({a}^{-1})}^{-1}=e$$ |

shows that it is also the left inverse of $a$. Using this result, we then can write

$$ea=(a{a}^{-1})a=a({a}^{-1}a)=ae=a,$$ |

whence $e$ is a left identity element, too.

Title | redundancy of two-sidedness in definition of group |
---|---|

Canonical name | RedundancyOfTwosidednessInDefinitionOfGroup |

Date of creation | 2015-01-20 17:28:03 |

Last modified on | 2015-01-20 17:28:03 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 3 |

Author | pahio (2872) |

Entry type | Definition |