Unique x for all g in G such that ?

In summary: those "fun facts" we learned about factoring integers into primes in grade school, turn out to be useful after all.
  • #1
Poirot1
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0
Let G be a group, |G|=n and m an integer such that gcd(m,n)=1.

(i) show that implies

(ii)Hence show that for all g in G there is a unique x such that

(i) there exist a, b such that am+bn=1 so that .

Hence ok?

(ii) (i) shows uniqueness. Not sure about existence. Cheers.
 
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  • #2
Poirot said:
Let G be a group, |G|=n and m an integer such that gcd(m,n)=1.

(i) show that implies

(ii)Hence show that for all g in G there is a unique x such that

(i) there exist a, b such that am+bn=1 so that .

Hence ok?

(ii) (i) shows uniqueness. Not sure about existence. Cheers.
To show existence in (ii) define as . We know that is an injective map. Since is finite, is also surjective. Hence...
 
  • #3
Very clever. I know in part (i) that am+bn=1 -> mod(n), but I've attempted to derive this and have failed.
 
  • #4
Poirot said:
Very clever. I know in part (i) that am+bn=1 -> mod(n), but I've attempted to derive this and have failed.
Are you asking why is it true that such that ??
 
  • #5
caffeinemachine said:
Are you asking why is it true that such that ??

Actually forget about it. am = 1 (mod n) because am - 1 is a multiple of n (which is what I was asking)
 
  • #6
Poirot said:
Actually forget about it. am = 1 (mod n) because am - 1 is a multiple of n (which is what I was asking)
Okay.
 
  • #7
you can also write it this way:

since











as an example of how this works, suppose

and we have .

if , then since has no elements of order 2.

if , then , and , so must be .

if , then and but , so .

what are the a and b in this case?

clearly 1 = (-1)2 + (1)3

so

the map given by is:



<--clearly bijective (in this case it's just the inversion map).

those "fun facts" we learned about factoring integers into primes in grade school, turn out to be useful after all.
 

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