What is the Non-Simplest Group of Order mp?

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SUMMARY

The discussion centers on proving that a group G of order mp, where p is prime and 1 PREREQUISITES

  • Understanding of group theory, specifically the concepts of group order and simplicity.
  • Familiarity with Cauchy's Theorem and its implications for subgroup existence.
  • Knowledge of group actions and cosets.
  • Basic understanding of homomorphisms and symmetric groups, particularly S_m.
NEXT STEPS
  • Study the implications of Cauchy's Theorem in various group contexts.
  • Learn about group actions and their applications in proving subgroup properties.
  • Explore the structure of symmetric groups, focusing on S_m and its properties.
  • Investigate normal subgroups and their significance in group theory.
USEFUL FOR

This discussion is beneficial for students of abstract algebra, particularly those studying group theory, as well as educators and researchers looking to deepen their understanding of group simplicity and subgroup structures.

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Homework Statement


Let G be a group with |G|= mp where p is prime and 1<m<p. Prove that G is not simple


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The Attempt at a Solution


I have proven the existence of a subgroup H that has order p(via Cauchy's Theorem), but I don't know how to use the representation of G on cosets of H or another method to somehow deduce that H is a normal subgroup of G thus forcing G to be not simple.
 
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Let G act on the cosets of H by left translation. This induces a homomorphism f:G->S_[G:H]=S_m (why?). Since kerf sits in H (why?), we can consider [H:kerf]. Since |H|=p, it follows that either [H:kerf]=1 or [H:kerf]=p (why?). If it's the former, we're done (why?). So suppose that [H:kerf]=p and deduce a contradiction.
 

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