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Determine Whether this is a Group

  • Thread starter cwatki14
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  • #1
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I am trying to answer the following question:
Determine whether (N,*) is a group or not. N={0,1,2,3,...} and
n*m= n+m if n,m even, -(n+m)-2 if n,m odd, and m-n-1 if n odd, m even.

I know that the properties of a group are associativity, existence of an identity, and inversiblity. If I consider the case where n,m are both even, clearly the third condition is not satisfied since there does not exist an additive inverse in N. Is this enough to conclude that it is not a group, or do I need to consider the other cases?
 

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  • #2
tiny-tim
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hi cwatki14! :smile:
Determine whether (N,*) is a group or not.

… the third condition is not satisfied … Is this enough to conclude that it is not a group, or do I need to consider the other cases?
no, disproving any essential condition is enough

(but isn't there also a condition that an identity has to exist, ie n*0 = n ?)

(and anyway N doesn't include negative numbers, so how do the definitions work? :confused:)
 
  • #3
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I think the identity element does exist. The problem statement defines N as including 0 as an element. As far as including the negative numbers, obviously these are not included. I think the definitions just define the operations performed on N in different cases of even, odd properties of 2 elements in N, right? I don't think it is claiming that negative numbers are a part of the set.
 
  • #4
tiny-tim
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but then how can they define …
-(n+m)-2 if n,m odd
… when -(n+m)-2 must be negative, and so can't be in N ? :confused:

(and what is n*0 or 0*n if n is odd? :confused:)
 

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