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normal subgroup problem


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Actually both. I can prove it by using axioms & definitions, 1st by showing that H is a sgp & then its normal. But the question holds only 2 marks. So I think there is a shorter way of doing it as well. I want help regarding that. Thank you.

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  • 1 year later...
On 12/20/2018 at 9:10 AM, taeto said:

I suspect you are supposed to see that G is abelian. That certainly helps for showing normality.

Good tip.+1. Is this homework, @Prasant36?

Edit: You also need Abelian character for showing closure

Edited by joigus
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16 hours ago, joigus said:

 

Edit: You also need Abelian character for showing closure

in mathematics, closure can correspond many things.may I ask:  which type of closure do you meantion here?

Edited by ahmet
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1 minute ago, ahmet said:

Algebraic closure,

Bingo!!

"Is a group" refers to algebraic properties.

\[g_{1},g_{2}\in H\Rightarrow g_{1}g_{2}\in H\]

We're not talking topological groups. (I'm not aware that anybody mentioned a basis of neighbourhoods). Welcome to page 1.

Neither have I read anything about a metric space.

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1 hour ago, ahmet said:

in mathematics, closure can correspond many things.may I ask:  which type of closure do you meantion here?

Also, by "normal" (in this context) I understand:

\[gHg^{-1}\subseteq H\]

Not "perpendicular". Any more questions?

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Just now, joigus said:

Also, by "normal" (in this context) I understand:

 

gHg1H

 

Not "perpendicular". Any more questions?

no (more) questions ,I just tried to understand what you meant 

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Just now, ahmet said:

I think almost all parts of mathematics have intersections (even topology and functional analysis with algebra)

....

They do. I know, and you know. And I know you know. And you know I know you know.

Can we stick to the topic, please? ;)  It's algebra. Group theory. That's why we are @ 

1 minute ago, ahmet said:

no (more) questions ,I just tried to understand what you meant 

Ah, OK. I'm sorry if I misunderstood your question in any sense.

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