GRE0568 Problems 43 and 49

Forum for the GRE subject test in mathematics.
EmLasker
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Joined: Sun Oct 05, 2008 1:41 am

GRE0568 Problems 43 and 49

Postby EmLasker » Sun Oct 12, 2008 11:02 pm

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

JcraigMSU
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Postby JcraigMSU » Mon Oct 13, 2008 1:57 am

For 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.

EmLasker
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Postby EmLasker » Mon Oct 13, 2008 2:18 am

Thank you for the reply...

Just to be sure I understand, the additive abelian groups of order 8 such that x+x = 0 would be:

Z4 + Z2 and Z2 + Z2 + Z2

and, the additive abelian groups of order 10 such that x+x+x+x+x=0 would be:

just Z5 + Z2


Thanks again,
Peter

CoCoA
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Postby CoCoA » Mon Oct 13, 2008 3:54 am

43: since z is a 5th root of unity, z^5=1, z^6=z, etc. giving
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.

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lime
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Postby lime » Mon Oct 13, 2008 9:21 am

Just to be sure I understand, the additive abelian groups of order 8 such that x+x = 0 would be:
Z4 + Z2

Incorrect. Counterexample: (1,1) + (1,1) = (2,0)

EmLasker
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Postby EmLasker » Mon Oct 13, 2008 12:18 pm

Cocoa,
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...

moo5003
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Postby moo5003 » Mon Oct 13, 2008 3:23 pm

Sounds right, Just remember that the LCM of the group orders (Call it K) is the lowest possible power such that x^K = 0 in an additive abelian group such as these.

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.




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