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Abstract Alegbra I

Posted: Tue Sep 20, 2011 11:02 am
by YKM
Dear all, I never really took abstract alegbra, so your help is highly appreciated.

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.

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.

The answer is E.

The answer is A.

The answer is B.

Re: Abstract Alegbra I

Posted: Tue Sep 20, 2011 1:41 pm
by blitzer6266
59) ... _theory%29

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

49) ... 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:


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.

Re: Abstract Alegbra I

Posted: Tue Sep 20, 2011 9:29 pm
by YKM
thank you very much.