Proving Normality of Subgroups in Cyclic Groups

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SUMMARY

The discussion focuses on proving that if all cyclic subgroups of a group G are normal, then all subgroups of G are also normal. The participant correctly identifies that all cyclic groups are abelian, which implies that G itself is abelian since it is a subgroup of itself. The conclusion drawn is that every subgroup of an abelian group is normal, thus confirming the statement. The suggestion to use proof by contradiction is also noted as a valid approach to explore the implications of a non-normal subgroup.

PREREQUISITES
  • Understanding of group theory concepts, specifically cyclic groups and normal subgroups.
  • Familiarity with the properties of abelian groups.
  • Knowledge of proof techniques, including proof by contradiction.
  • Basic comprehension of abstract algebra terminology.
NEXT STEPS
  • Study the properties of abelian groups in detail.
  • Learn about normal subgroups and their significance in group theory.
  • Explore proof techniques in abstract algebra, focusing on contradiction and direct proofs.
  • Investigate examples of cyclic groups and their subgroups to solidify understanding.
USEFUL FOR

Students of abstract algebra, mathematicians interested in group theory, and anyone seeking to deepen their understanding of subgroup properties and normality in cyclic groups.

Kuzu
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I'm taking this course "abstract algebra" at university and I've been given some homework questions. I was able to solve all of them but one. And it would be great if anyone could help me with this.

The question is like this:
"If all cyclic subgroups of G are normal, then show that all subgroups of G are normal"


as I know all cyclic groups are abelian, and G itself is a subgroup of G so it is cyclic and abelian. Also I know that every subgroup of an abelian group is normal. So I didn't even understand the question completely.

How can I solve this?
 
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Try a proof by contradiction. Take H to be a subgroup of G and suppose H is not normal. What does that tell you?
 

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