Correspondence Theorem for Normal Subgroups in Groups of Order 168

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SUMMARY

The discussion centers on proving that a group G of order 168 with a normal subgroup H of order 4 must also have a normal subgroup of order 28. The key steps involve applying the correspondence theorem, which relates the subgroups of G to the subgroups of the quotient group G/H. Given that |G/H| equals 42, the unique Sylow 7-subgroup K in G/N implies the existence of a normal subgroup of order 28. The correspondence theorem is essential for establishing this relationship between the subgroup orders.

PREREQUISITES
  • Understanding of group theory concepts, specifically normal subgroups
  • Familiarity with Sylow theorems and their applications
  • Knowledge of the correspondence theorem in group theory
  • Basic principles of group order and quotient groups
NEXT STEPS
  • Study the correspondence theorem in detail and its implications for subgroup structures
  • Explore Sylow's theorems, particularly in the context of groups of order 168
  • Investigate the properties of normal subgroups and their significance in group theory
  • Practice problems involving the application of theorems to find subgroup orders in finite groups
USEFUL FOR

Students of abstract algebra, particularly those studying group theory, as well as educators and researchers focusing on finite group structures and subgroup properties.

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Homework Statement



Show that if G is a group of order 168 that has a normal subgroup of order 4, then G has a normal subgroup of order 28.


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The Attempt at a Solution


Let H be a normal subgroup of order 4. Then |G/H|=42=2*3*7, so then G?N has a unique, and therefore normal Sylow 7-subgroup, let's call it K.
I was told to use the correspondence theorem, but I am not sure where it works in here. any ideas?
 
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Why don't you start by stating what the correspondence theorem says? The result you are seeking is an immediate consequence.
 

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