These two are giving me a bit of trouble...

I think I'm missing the point on this one, I know I could right everything out in terms of sin and cos, but there's got to be an easier and quicker way...

43. If z=exp(2*pi*i/5), then find 1 + z + z^2 + z^3 + 5z^4 + 4z^5 + 4z^6 + 4z^7 + 4z^8 + 5z^9.

I'm not even sure how to approach this one...

49. Up to isomorphism, how many additive abelian groups G of order 16 have the property that x + x + x + x = 0 for each x in G?

Peter

## GRE0568 Problems 43 and 49

x+x+x+x=0 in additive groups implies x^4 = 0.

For all x in G implies the order of every element in G must be 1, 2 or 4.

Using the Fundamental Theorem of Finite Abelian Groups, a group of order 16 breaks down and gives us possibilities that fulfill our element order property:

Z4 + Z4, Z4+Z2+Z2, Z2+Z2+Z2+Z2 = three possibilities.

1 + z + z^2 + z^3 + 5z^4 + 4z^5 + 4z^6 + 4z^7 + 4z^8 + 5z^9 =

5 + 5z + 5z^2 + 5z^3 + 10z^4

Then since z \neq 1, you should remember 1 + z + z^2 + z^3 + z^4 = 0

[since 0 = 1 - z^5 = (1-z)(1+z+z^2+z^3+z^4) ]

giving 5z^4

the rest you just work out.

Thank you for your reply, that was very helpful.

Lime,

is it that I shouldn't have Z4 to begin with, as x+x=0 implies that all subgroups are of order 1 or 2? Thus the only possible group for the problem I posed would be Z2+Z2+Z2? And, if in that case x+x+x+x=0 both Z4+Z2 and Z2+Z2+Z2 would work...

Thus if you want to find order 8 groups such that x+x+x+x=0 just find all groups with only z_2 and z_4.

Or in the other example for x+x=0 only use z_2.