Title: 群 (The Swarm (Schätzing novel))
Section: 正式定義
Subsection: 等價的定義

上面關於單位元和逆元素的部分也可以改為：因為不管原來的淡紫色定義，還是淡黃色的替代性定義，配上結合律都會等價於以下的定義： 以下是證明：若二元運算                     ∘        :        G        ×        G        →        G              {\displaystyle \circ :G\times G\to G}   滿足結合律，且                               e                      l                          ∈        G              {\displaystyle e_{l}\in G}   對任意                     g        ∈        G              {\displaystyle g\in G}    有「                              e                      l                          ∘        g        =        g              {\displaystyle e_{l}\circ g=g}   」(left-1)「存在某                               γ                      l                          ∈        G              {\displaystyle \gamma _{l}\in G}   使                               γ                      l                          ∘        g        =                  e                      l                                {\displaystyle \gamma _{l}\circ g=e_{l}}   」(left-2)因為                     ∘              {\displaystyle \circ }   是二元運算，                     g        ∘        γ        ∈        G              {\displaystyle g\circ \gamma \in G}   。所以根據(left-2)，存在                     Γ        ∈        G              {\displaystyle \Gamma \in G}   滿足                    Γ        ∘        (        g        ∘                  γ                      l                          )        =                  e                      l                                {\displaystyle \Gamma \circ (g\circ \gamma _{l})=e_{l}}  這樣根據結合律和(left-1)與(left-2)有                                                                                          e                                      l                                                                                              =                Γ                ∘                (                g                ∘                                  γ                                      l                                                  )                                                                                                  =                Γ                ∘                [                                g                ∘                (                                  e                                      l                                                  ∘                                  γ                                      l                                                  )                                ]                                                                                                  =                Γ                ∘                {                                g                ∘                [                                (                                  γ                                      l                                                  ∘                g                )                ∘                                  γ                                      l                                                                  ]                                }                                                                                                  =                Γ                ∘                {                                g                ∘                [                                                  γ                                      l                                                  ∘                (                g                ∘                                  γ                                      l                                                  )                                ]                                }                                                                                                  =                Γ                ∘                [                                (                g                ∘                                  γ                                      l                                                  )                ∘                (                g                ∘                                  γ                                      l                                                  )                                ]                                                                                                  =                [                                Γ                ∘                (                g                ∘                                  γ                                      l                                                  )                                ]                ∘                (                g                ∘                                  γ                                      l                                                  )                                                                                                  =                                  e                                      l                                                  ∘                (                g                ∘                                  γ                                      l                                                  )                                                                                                  =                g                ∘                                  γ                                      l                                                                                            {\displaystyle {\begin{aligned}e_{l}&=\Gamma \circ (g\circ \gamma _{l})\\&=\Gamma \circ [\,g\circ (e_{l}\circ \gamma _{l})\,]\\&=\Gamma \circ \{\,g\circ [\,(\gamma _{l}\circ g)\circ \gamma _{l}\,]\,\}\\&=\Gamma \circ \{\,g\circ [\,\gamma _{l}\circ (g\circ \gamma _{l})\,]\,\}\\&=\Gamma \circ [\,(g\circ \gamma _{l})\circ (g\circ \gamma _{l})\,]\\&=[\,\Gamma \circ (g\circ \gamma _{l})\,]\circ (g\circ \gamma _{l})\\&=e_{l}\circ (g\circ \gamma _{l})\\&=g\circ \gamma _{l}\end{aligned}}}  所以左逆元必為右逆元，這樣根據結合律有                                                                        g                ∘                                  e                                      l                                                                                              =                g                ∘                (                                  γ                                      l                                                  ∘                g                )                                                                                                  =                (                g                ∘                                  γ                                      l                                                  )                ∘                g                                                                                                  =                                  e                                      l                                                  ∘                g                =                g                                                          {\displaystyle {\begin{aligned}g\circ e_{l}&=g\circ (\gamma _{l}\circ g)\\&=(g\circ \gamma _{l})\circ g\\&=e_{l}\circ g=g\end{aligned}}}  所以左單位元必為右單位元。類似地，若二元運算                     ∘        :        G        ×        G        →        G              {\displaystyle \circ :G\times G\to G}   滿足結合律，且                               e                      r                          ∈        G              {\displaystyle e_{r}\in G}   對所有                     g        ∈        G              {\displaystyle g\in G}   有 「                     g        ∘                  e                      r                          =        g              {\displaystyle g\circ e_{r}=g}   」(right-1)「存在某                               γ                      r                          ∈        G              {\displaystyle \gamma _{r}\in G}   ，使                     g        ∘                  γ                      r                          =                  e                      r                                {\displaystyle g\circ \gamma _{r}=e_{r}}   」(right-2)因為                               γ                      r                          ∘        g        ∈        G              {\displaystyle \gamma _{r}\circ g\in G}   ，所以根據(right-2)存在                                           Γ            ¯                          ∈        G              {\displaystyle {\overline {\Gamma }}\in G}   滿足                    (                  γ                      r                          ∘        g        )        ∘                              Γ            ¯                          =                  e                      r                                {\displaystyle (\gamma _{r}\circ g)\circ {\overline {\Gamma }}=e_{r}}  所以根據結合律、(right-1)和(right-2)有                                                                                          e                                      r                                                                                              =                (                                  γ                                      r                                                  ∘                g                )                ∘                                                      Γ                    ¯                                                                                                                                    =                [                                (                                  γ                                      r                                                  ∘                                  e                                      r                                                  )                ∘                g                                ]                ∘                                                      Γ                    ¯                                                                                                                                    =                {                                [                                                  γ                                      r                                                  ∘                (                g                ∘                                  γ                                      r                                                  )                                ]                ∘                g                                }                ∘                                                      Γ                    ¯                                                                                                                                    =                {                                [                                (                                  γ                                      r                                                  ∘                g                )                ∘                                  γ                                      r                                                                  ]                ∘                g                                }                ∘                                                      Γ                    ¯                                                                                                                                    =                [                                (                                  γ                                      r                                                  ∘                g                )                ∘                (                                  γ                                      r                                                  ∘                g                )                                ]                ∘                                                      Γ                    ¯                                                                                                                                    =                (                                  γ                                      r                                                  ∘                g                )                ∘                [                                (                                  γ                                      r                                                  ∘                g                )                ∘                                                      Γ                    ¯                                                                  ]                                                                                                  =                (                                  γ                                      r                                                  ∘                g                )                ∘                                  e                                      r                                                                                                                                    =                                  γ                                      r                                                  ∘                g                                                          {\displaystyle {\begin{aligned}e_{r}&=(\gamma _{r}\circ g)\circ {\overline {\Gamma }}\\&=[\,(\gamma _{r}\circ e_{r})\circ g\,]\circ {\overline {\Gamma }}\\&=\{\,[\,\gamma _{r}\circ (g\circ \gamma _{r})\,]\circ g\,\}\circ {\overline {\Gamma }}\\&=\{\,[\,(\gamma _{r}\circ g)\circ \gamma _{r}\,]\circ g\,\}\circ {\overline {\Gamma }}\\&=[\,(\gamma _{r}\circ g)\circ (\gamma _{r}\circ g)\,]\circ {\overline {\Gamma }}\\&=(\gamma _{r}\circ g)\circ [\,(\gamma _{r}\circ g)\circ {\overline {\Gamma }}\,]\\&=(\gamma _{r}\circ g)\circ e_{r}\\&=\gamma _{r}\circ g\end{aligned}}}  所以右逆元也為左逆元。這樣根據結合律有                                                                                          e                                      r                                                  ∘                g                                                            =                (                g                ∘                                  γ                                      r                                                  )                ∘                g                                                                                                  =                g                ∘                (                                  γ                                      r                                                  ∘                g                )                                                                                                  =                g                ∘                                  e                                      r                                                  =                g                                                          {\displaystyle {\begin{aligned}e_{r}\circ g&=(g\circ \gamma _{r})\circ g\\&=g\circ (\gamma _{r}\circ g)\\&=g\circ e_{r}=g\end{aligned}}}  所以右單位元必為左單位元。再考慮到淡藍色是要求                     e        ∈        G              {\displaystyle e\in G}   同時為左右單位元且存在左右逆元，這樣就證明以上三種定義在有結合律的前提下等價。