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Lagrange’s Theorem

  • M. A. Armstrong
Part of the Undergraduate Texts in Mathematics book series (UTM)

Abstract

Consider a finite group G together with a subgroup H of G. Are the orders of H and G related in any way? Assuming H is not all of G, choose an element g 1 from GH, and multiply every element of H on the left by g 1 to form the set
$${g_1}H = \left\{ {{g_1}h|h \in H} \right\}$$
We claim that g 1 H has the same size as H and is disjoint from H. The first assertion follows because the correspondence hg 1 h from H to g 1 H can be inverted (just multiply every element of g 1 H on the left by \(g_1^{ - 1}\)) and is therefore a bijection. For the second, suppose x lies in both H and g 1 H. Then there is an element h 1H such that x = g 1 h 1. But this gives \({g_1} = xh_1^{ - 1}\), which contradicts our initial choice of g 1 outside H.

Keywords

Finite Group Identity Element Initial Choice Finite Subgroup Finite Abelian Group 
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

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Copyright information

© Springer Science+Business Media New York 1988

Authors and Affiliations

  • M. A. Armstrong
    • 1
  1. 1.Department of Mathematical SciencesUniversity of DurhamDurhamEngland

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