Abstract Alegbra I

Forum for the GRE subject test in mathematics.
YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

Abstract Alegbra I

Postby YKM » Tue Sep 20, 2011 11:02 am

Dear all, I never really took abstract alegbra, so your help is highly appreciated.

9367:
#59
Two subgroups H and K of a group G have orders 12 and 30, respectively. Which of the following could NOT be the order of the subgroup of G generated by H and K?

The answer is A: 30.


0568:
#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?

The answer is D: 3.

9367
#55
The answer is E.

9768
#59
The answer is A.

9768
#19
The answer is B.

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: Abstract Alegbra I

Postby blitzer6266 » Tue Sep 20, 2011 1:41 pm

59) http://en.wikipedia.org/wiki/Lagrange%2 ... _theory%29

12 does not divide 30 so 30 cannot be the order of the group.

49) http://en.wikipedia.org/wiki/Fundamenta ... sification

Basically you count the different ways of writing 16 as factors where order doesn't matter and none of those factors are bigger than 4. The three are:

2*2*2*2
2*2*4
4*4

55) Any subgroup of the additive group of integers is in the form {z | z=nm where m is an integer} where n is a fixed integer. For example if 3=n, the subgroup would be {...-6, -3, 0, 3, 6, 9...}. From here, you can see that if p=n, we have the elements p, pq, and p^q in the subgroup because they are divisible by p while p+q and q^p are not divisible by p.

59) If there are two distinct elements, then two of those elements in the set must be equal. We try all three cases.
If x^5=x^3, then by applying x^-3 to both sides, we get x^2=Id. But the order of an element must divide the order of the group, so this says x=id. Then all of the three elements are equal.

If x^5=x^9, then x^4=Id, so x=Id again.

But if x^9=x^3, then x^6=Id, so we could have x^3=Id where x is not equal to the identity.

Then x^13n= x^n*x^12n= x^n*(x^3)^4n= x^n. But x has order 3 so the three elements are x, x^2, and x^3=id.

19) For any group, if a and b are distinct, you can't have c*a= c*b (because if you applied c^-1, you'd get a=b).

Table 2 says c*c= c*d and Table 3 says d*a=d*c. Therefore, they are not groups.

For Table 1, if you replace a with 0, b with 1, c with 2, and d with 3, this is just the additive group of integers modulo 4.

YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

Re: Abstract Alegbra I

Postby YKM » Tue Sep 20, 2011 9:29 pm

thank you very much.




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