# this is used to prove cauchy's theorem(james mcKay)

• Jan 11th 2011, 03:34 AM
abhishekkgp
this is used to prove cauchy's theorem(james mcKay)
Let G be finite group and let p be a prime dividing |G|. Let S denote the set of p-tuples of elements of G the product of whose coordinate is 1:

S={(x1,x2,x3,...,xp)|and x1*x2*x3*...*xp=1}
where x1,x2,x3... belong to G. and '*' represents the binary operation under which G is a group. Note that x1,x2,x3... need not be distinct.

PROVE THAT: S has |G|^(p-1) elements.
• Jan 11th 2011, 04:01 AM
Swlabr
Quote:

Originally Posted by abhishekkgp
Let G be finite group and let p be a prime dividing |G|. Let S denote the set of p-tuples of elements of G the product of whose coordinate is 1:

S={(x1,x2,x3,...,xp)|and x1*x2*x3*...*xp=1}
where x1,x2,x3... belong to G. and '*' represents the binary operation under which G is a group. Note that x1,x2,x3... need not be distinct.

PROVE THAT: S has |G|^(p-1) elements.

you have |G| choices for the first p-1 elements. How many choice does this leave for the last element?
• Jan 11th 2011, 06:31 AM
abhishekkgp
Quote:

Originally Posted by Swlabr
you have |G| choices for the first p-1 elements. How many choice does this leave for the last element?

:)
thanks.. that helped.. although now it seem really simple.

3987^{12}+4365^{12}=4472^{12} ?? isnt this wrong? doesnt satisfy divisibility by 3(as well as invalidates FLT)
• Jan 11th 2011, 06:36 AM
Swlabr
Yes, it is wrong. However, they match for the first 10 digits, so at a glance...
• Jan 11th 2011, 06:43 AM
abhishekkgp
whoa!!