Remark 1 on Groups

Let $(G, \circ)$ be a group. The following holds:

i)	If $a^{'}$ is the inverse of $a \in G$ , then $a \circ a^{'} = e$ holds. This means that if the inverse element $a^{'}$ of $a$ is combined with the element $a$ from the right through the operation $\circ$ , it also results in the neutral element $e$ . In G3, only the existence of an inverse element $a^{'}$ is described, which, when combined with the element $a$ from the left through the operation $\circ$ , results in the neutral element. Thus, it is described here that the order of the two elements $a$ and $a^{'}$ in their combination does not matter, and $a \circ a^{'} = a^{'} \circ a = e$ holds.
ii)	Let $e \in G$ be a neutral element. Then $a \circ e = a$ holds for all $a \in G$ . This means that if the neutral element $e \in G$ is combined with the element $a$ from the right through the operation $\circ$ , it also results in the element $a$ . In G2, only the existence of a neutral element $e$ is described, which, when combined with the element $a$ from the left through the operation $\circ$ , again results in the element $a$ . Thus, it is described here that the order in the combination of an element $a$ with the neutral element $e$ does not matter, and $a \circ e = e \circ a = a$ holds.
iii)	There is exactly one neutral element $e \in G$ .
iv)	For each element, there is exactly one inverse element $a^{'}$ , which is also denoted by $a^{- 1}$ . Here, the uniqueness of the inverse element is described.

Beweis

i)	According to the definition of groups G3, for $a^{'} \in G$ , there exists an inverse element $(a^{'})^{'} = a^{″} \in G$ with $a^{″} \circ a^{'} = e$ . It follows that \begin{align} a \circ a' &= e \circ (a \circ a') \ &\text{here the neutral element } e \text{ is replaced by } a'' \circ a' \ &= (a'' \circ a') \circ (a \circ a') \ &\text{according to G1, it does not matter which elements are combined first} \ &\overset{\text{G1}}{=} a'' \circ ((a' \circ a) \circ a') \ &\text{Since } a' \circ a \text{ results in the neutral element } e, \text{ it can be replaced} \ &\overset{\text{G3}}{=} a'' \circ (e \circ a') \ &\text{The combination with the neutral element } e \text{ does not change the element } a' \ &\overset{\text{G2}}{=} a'' \circ a' \ &\text{since } a'' \text{ is the inverse element to } a', \text{ this results in the neutral element } e \ &= e. \end{align}
ii)	It holds that \begin{align} a \circ e &\overset{\text{G3}}{=} a \circ (a' \circ a) \ &\text{the neutral element } e \text{ is replaced by } a' \circ a \ &\overset{\text{G1}}{=} (a \circ a') \circ a \ &\text{according to G1, it does not matter which elements are combined first} \ &\overset{(i)}{=} e \circ a \ &\text{due to (i), since } a \circ a' = e, \text{ the term } a \circ a' \text{ can be replaced by } e \ &\overset{\text{G2}}{=} a. \end{align}
iii)	Let $e^{'}$ be another neutral element besides the neutral element $e$ described in G2. It will now be shown that this additional neutral element $e^{'}$ must be the same as the neutral element $e$ . If the additional neutral element $e^{'}$ is combined with the neutral element $e$ , it follows that: \begin{align} e' &\overset{\text{G2}}{=} e' \circ e \ &\overset{(ii)}{=} e \end{align} because the combination with a neutral element according to (i) does not change the element it is combined with (even if the element it is combined with is the neutral element).
iv)	Let $a^{'}$ be the inverse element to $a$ according to G3. This means $a^{'} \circ a = a \circ a^{'} = e$ holds. Let $a^{}$ be another inverse element to $a$ . This means $a^{} \circ a = a \circ a^{} = e$ holds. Now it will be shown that $a^{'} = a^{}$ holds. \begin{align} a' &= a' \circ e \ &\text{since } a^{} \text{ is an inverse element to } a, \text{ } e \text{ can be replaced by } a \circ a^{} \ &= a' \circ (a \circ a^{}) \ &= (a' \circ a) \circ a^{} \ &= e \circ a^{} \ &= a^{} \end{align}