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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Induced homomorphism (quotient group) ： ウィキペディア英語版 Induced homomorphism (quotient group) In mathematics, specifically group theory, a homomorphism $\backslash phi:\backslash ,G\backslash to\; H$ from one group $G$ to another group $H$ may give rise to an induced homomorphism $\backslash tilde:\backslash ,\backslash tilde\backslash to\backslash tilde$ between groups $\backslash tilde$ and $\backslash tilde$ which are associated canonically with $G$ and $H$, provided the original homomorphism $\backslash phi$ satisfies suitable conditions. One of the most important situations, with lots of useful applications, arises when normal subgroups $U\backslash triangleleft\; G$ and $V\backslash triangleleft\; H$ are given and we search for necessary and sufficient conditions for the existence of an induced homomorphism $\backslash tilde:\backslash ,G/U\backslash to\; H/V$ between the corresponding quotient groups $G/U$ and $H/V$ which is connected with $\backslash phi$ in a natural way. ==Image, pre-image and kernel== Throughout this article, let $\backslash phi:\backslash ,G\backslash to\; H$ be a homomorphism from a source group (domain) $G$ to a target group (codomain)$H$. First, we recall some basic facts concerning the image and pre-image of normal subgroups and the kernel of the homomorphism $\backslash phi$, as given in Satz 40, p. 44, of Reiffen, Scheja and Vetter. 〔 Lemma. Suppose that $U\backslash le\; G$ and $V\backslash le\; H$ are subgroups, and $x,y\backslash in\; G$ are elements. :1. If $U\backslash triangleleft\; G$ is a normal subgroup of $G$, then its image $\backslash phi(U)\backslash triangleleft\backslash phi(G)$ is a normal subgroup of the (total) image $\backslash mathrm(\backslash phi)=\backslash phi(G)$. :2. If $V\backslash triangleleft\backslash phi(G)$ is a normal subgroup of the image $\backslash mathrm(\backslash phi)=\backslash phi(G)$, then the pre-image $\backslash phi^(V)\backslash triangleleft\; G$ is a normal subgroup of $G$. :3. In particular, the kernel $\backslash ker(\backslash phi)=\backslash phi^(1)\backslash triangleleft\; G$ of $\backslash phi$ is a normal subgroup of $G$. :4. If $\backslash phi(x)=\backslash phi(y)$, then there exists an element $k\backslash in\backslash ker(\backslash phi)$ such that $x=y\backslash cdot\; k$. :5. If $\backslash phi(x)\backslash in\backslash phi(U)$, then $x\backslash in\; U\backslash cdot\backslash ker(\backslash phi)$, i.e., the pre-image of the image satisfies $(1)\backslash qquad\; U\backslash le\backslash phi^(\backslash phi(U))=U\backslash cdot\backslash ker(\backslash phi)=\backslash ker(\backslash phi)\backslash cdot\; U$. :6. Conversely, the image of the pre-image is given by $(2)\backslash qquad\; \backslash phi(\backslash phi^(V))=\backslash phi(G)\backslash bigcap\; V\backslash le\; V$. The situation of the Lemma is visualized by Figure 1, where we briefly write $K:=\backslash ker(\backslash phi)=\backslash phi^(1)$ and $I:=\backslash mathrm(\backslash phi)=\backslash phi(G)$. Remark. Note that, in the first statement of the Lemma, we cannot conclude that $\backslash phi(U)\backslash triangleleft\; H$ is a normal subgroup of the target group $H$, and in the second statement of the Lemma, we need not require that $V\backslash triangleleft\; H$ is a normal subgroup of the target group $H$. For the proof click ''show'' on the right hand side. 1. If $U\backslash triangleleft\; G$, then $x^Ux=U$ for all $x\backslash in\; G$, and thus $\backslash phi(x)^\backslash phi(U)\backslash phi(x)=\backslash phi(x^Ux)=\backslash phi(U)$ for all $\backslash phi(x)\backslash in\backslash phi(G)$, i.e., $\backslash phi(U)\backslash triangleleft\backslash phi(G)$. 2. If $V\backslash triangleleft\backslash phi(G)$, then $(\backslash forall\; x\backslash in\; G)\backslash \; \backslash phi(x)^V\backslash phi(x)\backslash subseteq\; V$, that is, $(\backslash forall\; x\backslash in\; G\backslash ,\backslash forall\; v\backslash in\; V)\backslash \; \backslash phi(x)^v\backslash phi(x)\backslash in\; V$. In particular, we have $(\backslash forall\; x\backslash in\; G\backslash ,\backslash forall\; u\backslash in\backslash phi^(V))\backslash \; \backslash phi(x^ux)=\backslash phi(x)^\backslash phi(u)\backslash phi(x)\backslash in\; V$, i.e., $(\backslash forall\; x\backslash in\; G)\backslash \; x^\backslash phi^(V)x\backslash subseteq\backslash phi^(V)$, and consequently $\backslash phi^(V)\backslash triangleleft\; G$. 3. To prove the claim for the kernel, we put $V:=1\backslash triangleleft\; G$ in the second statement. 4. If $\backslash phi(x)=\backslash phi(y)$, then $\backslash phi(y^x)=\backslash phi(y)^\backslash phi(x)=1$, and thus $y^x=k\backslash in\backslash ker(\backslash phi)$. 5. If $\backslash phi(x)\backslash in\backslash phi(U)$, then $(\backslash exists\; u\backslash in\; U)\backslash \; \backslash phi(x)=\backslash phi(u)$, and thus $(\backslash exists\; u\backslash in\; U\backslash ,\backslash exists\; k\backslash in\backslash ker(\backslash phi))\backslash \; x=u\backslash cdot\; k$, by the fourth statement. This shows $\backslash phi^(\backslash phi(U))\backslash subseteq\; U\backslash cdot\backslash ker(\backslash phi)$, and the opposite inclusion is obvious. Finally, since $\backslash ker(\backslash phi)$ is normal, we have $U\backslash cdot\backslash ker(\backslash phi)=\backslash ker(\backslash phi)\backslash cdot\; U$. 6. This is a consequence of the properties of the set mappings $\backslash phi^$ and $\backslash phi$ associated with the homomorphism $\backslash phi$. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Induced homomorphism (quotient group)」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.