What are the properties of normal subgroups in groups of prime order?

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SUMMARY

In the discussion, participants analyze the properties of normal subgroups within groups of prime order, specifically focusing on a group G of order p², where p is a prime number. It is established that any subgroup H of order p is normal in G, and that G must be abelian. The conversation also touches on the implications of Lagrange's theorem and the structure of cyclic groups, emphasizing the necessity of a nontrivial center in G to support these conclusions.

PREREQUISITES
  • Understanding of Lagrange's theorem in group theory
  • Familiarity with cyclic groups and their properties
  • Knowledge of the class equation in group theory
  • Concept of normal subgroups and their significance
NEXT STEPS
  • Study the implications of Lagrange's theorem on subgroup orders
  • Learn about the structure and properties of abelian groups
  • Explore the class equation and its applications in group theory
  • Investigate the concept of self-normalizing groups and their characteristics
USEFUL FOR

This discussion is beneficial for students of abstract algebra, particularly those studying group theory, as well as educators and researchers interested in the properties of normal subgroups and the structure of groups of prime order.

losiu99
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Hi, I've been vanquished by probably easy problems once again.

Homework Statement


1. Let G be a group of order p^2 (p prime number), and H its subgroup of order p. Show that H is normal. Prove G must be abelian.

2. If a group G has exactly one subgroup H of order k, prove H is normal in G.


Homework Equations


Lagrange theorem I think. Isomorphism theorems maybe?


The Attempt at a Solution


1. Obviously H is cyclic. If H is not normal, G cannot be abelian, hence all the elements are of order p, except for the neutral one. G is therefore divided into p+1 cyclic, disjoint (except for e) subgroups of order p. So far I haven't succeeded deriving a contradiction.

2. Normalizer is either H or the whole group. Perhaps some property of self-normalizing groups yields a contradiction?

Thank you very much for any hints.
 
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The second one isn't hard at all. Just think about the conjugate subgroups of H. The first one is a little harder. I'd start by using the class equation to show that G has a nontrivial center.
 

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