Lagrange’s Theorem

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


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.


Finite Group Identity Element Initial Choice Finite Subgroup Finite Abelian Group 
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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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