Prove H U K is Not a Subgroup of G | Groups and Subgroups

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Homework Help Overview

The discussion revolves around the properties of groups and subgroups, specifically focusing on the union of two subgroups H and K within a group G. The original poster attempts to prove that the union H ∪ K is not a subgroup of G by examining elements from H and K that do not overlap.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of finding elements h in H but not in K, and k in K but not in H. They discuss the failure of H ∪ K to meet the necessary properties of a group, particularly questioning which property is violated.

Discussion Status

The discussion is ongoing, with participants questioning the specific group property that H ∪ K fails to satisfy. There is an exchange of ideas about the nature of the elements involved and their membership in H ∪ K.

Contextual Notes

There is a focus on the definitions and properties of groups, particularly regarding the requirements for a set to be considered a subgroup. The participants are navigating through the implications of their findings without reaching a definitive conclusion.

xlalcciax
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1. Let G be a group containing subgroups H and K such that we can find an element h e H-K an an element k e K - H. Prove that h o k is not a subgroup of H U K. Deduce that H U K is not a subgroup of G.

I have proved that h o k is not in H U K but I don't know how to deduce that H U K is not a subgroup of G.
 
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If h \in H \setminus K, k \in K \setminus H, and you have proved that hk \notin H \cup K, that shows that H \cup K fails to satisfy one of the three basic properties required of a group. Which one?
 
ystael said:
If h \in H \setminus K, k \in K \setminus H, and you have proved that hk \notin H \cup K, that shows that H \cup K fails to satisfy one of the three basic properties required of a group. Which one?

Inverse element?
 
No, guess again...
 
ystael said:
If h \in H \setminus K, k \in K \setminus H, and you have proved that hk \notin H \cup K, that shows that H \cup K fails to satisfy one of the three basic properties required of a group. Which one?

xlalcciax said:
Inverse element?
Hi xlalcciax.

Are h, k ϵ H∪K ?
 

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