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.
Abstract Alegbra I

 Posts: 61
 Joined: Sun Apr 04, 2010 1:08 pm
Re: Abstract Alegbra I
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.
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.