Proving Core of a Group is Normal in G and Subset of H

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Homework Help Overview

The discussion revolves around proving properties of the core of a subgroup H within a group G, specifically that the core of H is normal in G and that it is a subset of H. The subject area is group theory.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the definition of the core of H and its implications for normality. There are attempts to prove that core(H) is a subgroup and to verify the normality condition through group operations.

Discussion Status

The discussion is active with participants sharing their reasoning and checking assumptions. Some guidance has been offered regarding the subgroup test and the conditions for normality, but no consensus has been reached on the proof steps.

Contextual Notes

Participants are navigating the definitions and properties of group elements and their interactions, with an emphasis on ensuring that all necessary conditions for subgroup and normality are met.

Punkyc7
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Let H be a subgroup of G and define the core of H as such
core H={g[itex]\in[/itex]G| g[itex]\in[/itex]aHa^-1 for all a[itex]\in[/itex]G}= [itex]\bigcap[/itex]{aHa^-1|a[itex]\in[/itex]G}
Prove that the core of H is normal in G and core H[itex]\subset[/itex]H.


I am having a hard time proving this because isn't the definition of core H basically saying the the core is normal?
 
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It is pretty close to tautological, but as I like to say if it's so obvious it should be easy to prove. You need to do two things:

1) Prove core(H) is actually a subgroup
2) Prove that for a∈G, acore(H)a-1=core(H)
 
Ok so would I say something like
If e is in core H becaus aea^-1=e
Let g and h^-1 be in the core then

ag^a-1(aha^-1)=agh^-1a^-1 ...is this right?
so its a a sub group by the one step subgroup test.
 
As long as you state that that's true for all a, that looks good to me
 
Ok thanks, I thought there might have been something I was missing
 

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