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        "quote": "> [!example] <a class=\"object-permalink\" href=\"#example-2-22\">例 2.22</a> \n>\n> 设$G$和$H$分别为两个具有恒等元$e_{G}$和$e_{H}$的群。可以在笛卡尔积$G \\times H$上定义一个合成律，即\n>\n> $$\n> (g, h)\\left(g^{\\prime}, h^{\\prime}\\right)=\\left(g g^{\\prime}, h h^{\\prime}\\right)\n> $$\n>\n> 这个乘积显然满足结合律，并且具有恒等元$\\left(e_{G}, e_{H}\\right)$。此外，每个元素都有唯一的逆元$(g, h)^{-1}=\\left(g^{-1}, h^{-1}\\right)$。因此，由于这个组合律，$G \\times H$是一个称为$G$和$H$的<u id=\"concept-2-direct-product\" class=\"concept-term\" data-concept-identity=\"urn:uuid:0267f40f-cd9e-43de-8086-9a9f751a68b8\" data-concept-route=\"/concept-identities#concept-0267f40f-cd9e-43de-8086-9a9f751a68b8\">直积(direct product)</u>的群。显然，群$G$与子群$\\left(G, e_{H}\\right)=\\left\\{\\left(g, e_{H}\\right) \\mid g \\in G\\right\\}$是同构的。后者是一个正规子群，因为\n>\n> $$\n> (a, b)\\left(G, e_{H}\\right)\\left(a^{-1}, b^{-1}\\right)=\\left(a G a^{-1}, b e_{H} b^{-1}\\right)=\\left(G, e_{H}\\right)\n> $$\n>\n> 通常将元素$\\left(G, e_{H}\\right)$的子群与群$G$认为是相同的。类似地，$H$被认为与正规子群$\\left(e_{G}, H\\right)$等同。",
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