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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 Artin transfer (group theory) ： ウィキペディア英語版 Artin transfer (group theory) In the mathematical field of group theory, an Artin transfer is a certain homomorphism from an arbitrary finite or infinite group to the commutator quotient group of a subgroup of finite index. Originally, such mappings arose as group theoretic counterparts of class extension homomorphisms of abelian extensions of algebraic number fields by applying Artin's reciprocity maps to ideal class groups and analyzing the resulting homomorphisms between quotients of Galois groups. However, independently of number theoretic applications, a partial order on the kernels and targets of Artin transfers has recently turned out to be compatible with parent-descendant relations between finite ''p''-groups (with a prime number ''p''), which can be visualized in descendant trees. Therefore, Artin transfers provide a valuable tool for the classification of finite ''p''-groups and for searching and identifying particular groups in descendant trees by looking for patterns defined by the kernels and targets of Artin transfers. These strategies of pattern recognition are useful in purely group theoretic context, as well as for applications in algebraic number theory concerning Galois groups of higher ''p''-class fields and Hilbert ''p''-class field towers. ==Transversals of a subgroup== Let $G$ be a group and $H\backslash le\; G$ be a subgroup of finite index $n=(G:H)\backslash ge\; 1$. Definitions. 〔 〕 :# A left transversal of $H$ in $G$ is an ordered system $(g\_1,\backslash ldots,g\_n)$ of representatives for the left cosets of $H$ in $G$ such that $G=\backslash dot\_^n\backslash ,g\_iH$ is a disjoint union. :# Similarly, a right transversal of $H$ in $G$ is an ordered system $(d\_1,\backslash ldots,d\_n)$ of representatives for the right cosets of $H$ in $G$ such that $G=\backslash dot\_^n\backslash ,Hd\_i$ is a disjoint union. Remark. For any transversal of $H$ in $G$, there exists a unique subscript $1\backslash le\; i\_0\backslash le\; n$ such that $g\_\backslash in\; H$, resp. $d\_\backslash in\; H$. Of course, this element with subscript $i\_0$ which represents the principal coset (i.e., the subgroup $H$ itself) may be, but need not be, replaced by the neutral element $1$. Lemma. 〔 :# If $G$ is non-abelian and $H$ is not a normal subgroup of $G$, then we can only say that the inverse elements $(g\_1^,\backslash ldots,g\_n^)$ of a left transversal $(g\_1,\backslash ldots,g\_n)$ form a right transversal of $H$ in $G$. :# However, if $H\backslash triangleleft\; G$ is a normal subgroup of $G$, then any left transversal is also a right transversal of $H$ in $G$. For the proof click ''show'' on the right hand side. :# Since the mapping $G\backslash to\; G,\backslash \; x\backslash mapsto\; x^$ is an involution, that is a bijection which is its own inverse, we see that $G=\backslash dot\_^n\backslash ,g\_iH$ implies $G=G^=\backslash dot\_^n\backslash ,(g\_iH)^=\backslash dot\_^n\backslash ,H^g\_i^=\backslash dot\_^n\backslash ,Hg\_i^$. :# For a normal subgroup $H\backslash triangleleft\; G$, we have $xH=Hx$ for each $x\backslash in\; G$. Let $\backslash phi:\backslash ,G\backslash to\; K$ be a group homomorphism and $(g\_1,\backslash ldots,g\_n)$ be a left transversal of a subgroup $H$ in $G$ with finite index $n=(G:H)\backslash ge\; 1$. We must check whether the image of this transversal under the homomorphism is again a transversal. Proposition. The following two conditions are equivalent. :# $(\backslash phi(g\_1),\backslash ldots,\backslash phi(g\_n))$ is a left transversal of the subgroup $\backslash phi(H)$ in the image $\backslash phi(G)$ with finite index $(\backslash phi(G):\backslash phi(H))=n$. :# $\backslash ker(\backslash phi)\backslash le\; H$. We emphasize this important equivalence in a formula: $(1)\backslash qquad\; \backslash phi(G)=\backslash dot\_^n\backslash ,\backslash phi(g\_i)\backslash phi(H)$ and $(\backslash phi(G):\backslash phi(H))=n\; \backslash quad\backslash Longleftrightarrow\backslash quad\; \backslash ker(\backslash phi)\backslash le\; H$. For the proof click ''show'' on the right hand side. By assumption, we have the disjoint left coset decomposition $G=\backslash dot\_^n\backslash ,g\_iH$ which comprises two statements simultaneously. Firstly, the group $G=\backslash bigcup\_^n\backslash ,g\_iH$ is a union of cosets, and secondly, any two distinct cosets have an empty intersection $g\_iH\backslash bigcap\; g\_jH=\backslash emptyset$, for $i\backslash ne\; j$. Due to the properties of the set mapping associated with $\backslash phi$, the homomorphism $\backslash phi$ maps the union to another union $\backslash phi(G)=\backslash phi(\backslash bigcup\_^n\backslash ,g\_iH)=\backslash bigcup\_^n\backslash ,\backslash phi(g\_iH)=\backslash bigcup\_^n\backslash ,\backslash phi(g\_i)\backslash phi(H)$, but weakens the equality for the intersection to a trivial inclusion $\backslash emptyset=\backslash phi(\backslash emptyset)=\backslash phi(g\_iH\backslash bigcap\; g\_jH)\backslash subseteq\backslash phi(g\_iH)\backslash bigcap\backslash phi(g\_jH)=\backslash phi(g\_i)\backslash phi(H)\backslash bigcap\backslash phi(g\_j)\backslash phi(H)$, for $i\backslash ne\; j$. To show that the images of the cosets remain disjoint we need the property $\backslash ker(\backslash phi)\backslash le\; H$ of the homomorphism $\backslash phi$. Suppose that $\backslash phi(g\_i)\backslash phi(H)\backslash bigcap\backslash phi(g\_j)\backslash phi(H)\backslash ne\backslash emptyset$ for some $1\backslash le\; i\backslash le\; j\backslash le\; n$, then we have $\backslash phi(g\_i)\backslash phi(h\_i)=\backslash phi(g\_j)\backslash phi(h\_j)$ for certain elements $h\_i,h\_j\backslash in\; H$. Multiplying by $\backslash phi(g\_j)^$ from the left and by $\backslash phi(h\_j)^$ from the right, we obtain $\backslash phi(g\_j^g\_ih\_ih\_j^)=\backslash phi(g\_j)^\backslash phi(g\_i)\backslash phi(h\_i)\backslash phi(h\_j)^=1$, that is, $g\_j^g\_ih\_ih\_j^\backslash in\backslash ker(\backslash phi)\backslash le\; H$. Since $h\_ih\_j^\backslash in\; H$, this implies $g\_j^g\_i\backslash in\; H$, resp. $g\_iH=g\_jH$, and thus $i=j$. Conversely, we use contraposition. If the kernel $\backslash ker(\backslash phi)$ of $\backslash phi$ is not contained in the subgroup $H$, then there exists an element $x\backslash in\; G\backslash setminus\; H$ such that $\backslash phi(x)=1$. But then the homomorphism $\backslash phi$ maps the disjoint cosets $xH\backslash bigcap\; 1\backslash cdot\; H=\backslash emptyset$ to equal cosets $\backslash phi(x)\backslash phi(H)\backslash bigcap\backslash phi(1)\backslash phi(H)=1\backslash cdot\backslash phi(H)\backslash bigcap\; 1\backslash cdot\backslash phi(H)=\backslash phi(H)$. 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「Artin transfer (group theory)」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.