Groups

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anandcu3

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Sep 30, 2011
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3 problems i am unable to solve. please help me out.

1) If every element of a group G is its own inverse. Prove that G is an abelian

2) Show That the set G={a^ n | n belogs to I} is an abelian under + mod 4

3)On q-{1} a set of rational numbers except 1. Define * as a*b = a+b-ab for all a,b belongs to q-{1}. Is this a group . justify

Please help me out.:(
 
Last edited:
anandcu3 said:
1) If every element of a group G is its own inverse. Prove that G is an abelian

2) Show That the set G={a to the power of n | n belogs to I} is an abelian under +4

3)On q-{1} a set of rational numbers except 1. Define * as a*b = a+b-ab for all a,b belongs to q-{1}. Is this a group . justify
Click to expand...
1) we know that \(\displaystyle (ab)(ab)=e\) so \(\displaystyle a(ba)b=e\).
There is just one more step to get \(\displaystyle ba=ab\) what is it?

2) I have no idea what that says.

3)
  • Can you show that 0 is the identity?
  • Is the operation associative?
  • Does each element have an inverse?
 
anandcu3 said:
3 problems i am unable to solve. please help me out.

1) If every element of a group G is its own inverse. Prove that G is an abelian

2) Show That the set G={a to the power of n | n belogs to I} is an abelian under +4

3)On q-{1} a set of rational numbers except 1. Define * as a*b = a+b-ab for all a,b belongs to q-{1}. Is this a group . justify

Please help me out.:(
Click to expand...

This post doesn't fit under this "Pre-Algebra" section at all.

To a moderator:

Please *move* this to an appropriate subforum.
 
Re:

@pka

Thanks Sir.

I understood the 3rd one.
But the 1st one i do not know how to start the problem itself.
I was unable to understand your explanation.
Please help me out.
:???:
 
anandcu3 said:
I understood the 3rd one.
But the 1st one i do not know how to start the problem itself.
I was unable to understand your explanation
Click to expand...
I do not know how unless you tell us what you do not understand. Do you know what it means to say every element is its own inverse? That is the reason for the first line in the proof.
 
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