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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*}