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In the mathematical subject of group theory, the Grushko theorem or the Grushko-Neumann theorem is a theorem stating that the rank (that is, the smallest cardinality of a generating set) of a free product of two groups is equal to the sum of the ranks of the two free factors. The theorem was first obtained in a 1940 article of Grushko〔I. A. Grushko, ''On the bases of a free product of groups'', Matematicheskii Sbornik, vol 8 (1940), pp. 169–182.〕 and then, independently, in a 1943 article of Neumann.〔B. H. Neumann. ''On the number of generators of a free product.'' Journal of the London Mathematical Society, vol 18, (1943), pp. 12–20.〕 ==Statement of the theorem== Let ''A'' and ''B'' be finitely generated groups and let ''A''∗''B'' be the free product of ''A'' and ''B''. Then :rank(''A''∗''B'') = rank(''A'') + rank(''B''). It is obvious that rank(''A''∗''B'') ≤ rank(''A'') + rank(''B'') since if X is a finite generating set of ''A'' and ''Y'' is a finite generating set of ''B'' then ''X''∪''Y'' is a generating set for ''A''∗''B'' and that |''X''∪''Y''|≤|''X''| + |''Y''|. The opposite inequality, rank(''A''∗''B'') ≥ rank(''A'') + rank(''B''), requires proof. There is a more precise version of Grushko's theorem in terms of Nielsen equivalence. It states that if ''M'' = (''g''1, ''g''2, ..., ''g''''n'') is an ''n''-tuple of elements of ''G'' = ''A''∗''B'' such that ''M'' generates ''G'', <''g''1, ''g''2, ..., ''g''''n''> = ''G'', then ''M'' is Nielsen equivalent in ''G'' to an ''n''-tuple of the form :''M = (''a''1, ..., ''a''''k'', ''b''1, ..., ''b''''n''−''k'') where ⊆''A'' is a generating set for ''A'' and where ⊆''B'' is a generating set for ''B''. In particular, rank(''A'') ≤ ''k'', rank(''B'') ≤ ''n'' − ''k'' and rank(''A'') + rank(''B'') ≤ ''k'' + (''n'' − ''k'') = ''n''. If one takes ''M'' to be the minimal generating tuple for ''G'', that is, with ''n'' = rank(''G''), this implies that rank(''A'') + rank(''B'') ≤ rank(''G''). Since the opposite inequality, rank(''G'') ≤ rank(''A'') + rank(''B''), is obvious, it follows that rank(''G'')=rank(''A'') + rank(''B''), as required. 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「Grushko theorem」の詳細全文を読む スポンサード リンク
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