Proposition: Simple Calculations Rules in a Group

In a group $(G,\ast)$, the following calculation rules follow immediately from the definition:

The inverse element of $e$ is $e$ itself, symbolically $e^{-1}=e$.
The equation $(x^{-1})^{-1}=x$ holds for all $x\in G$.
For all $a,b\in G$, the equation $a\ast x=b$ is uniquely solvable in $G$, and the solution is $x=a^{-1}\ast b$.
For all $a,b\in G$, the equation $y\ast a=b$ is uniquely solvable in $G$, and the solution is $y=b\ast a^{-1}.$
A group is cancellative, i.e. if $a\ast x=a\ast y$ then $x=y$ and if $x\ast a=y\ast a$ then $x=y.$
The equation $(x\ast y)^{-1}=y^{-1}\ast x^{-1}$ holds for all $x,y\in G$.
$x_1\ast x_2\ast x_3\ast \ldots\ast x_n:=(\ldots((x_1\ast x_2)\ast x_3)\ast \ldots\ast x_n$ holds for all $x_i\in G$.

If the group $(G,\ast)$ is Abelian, then it also follows that

8. $x_{k_1}\ast x_{k_2}\ast \ldots\ast x_{k_n}=x_{1}\ast x_{2}\ast \ldots\ast x_{n}$ holds for any permutation $k_1,\ldots,k_n$ of the indices $1,\ldots, n$.

Proofs: 1

Explanations: 1
Proofs: 2

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Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983