GR1268 Q12, 23, 29

Forum for the GRE subject test in mathematics.
mlbrown0
Posts: 1
Joined: Wed Oct 26, 2016 10:31 am

GR1268 Q12, 23, 29

Postby mlbrown0 » Wed Oct 26, 2016 11:41 am

Hi! I come from a physics background, so I wanted to check my understanding of some of the terminology around set theory questions. Please let me know if there are any mistakes!

Q12 asks which integers n within 3 \le n \le 11 has only one group of order n up to isomorphism. I understand this as "How many sets with n elements exist that cannot be mapped onto each other?"

Q23 begins "Let \left ( \mathbb{Z}_{10}, +, \cdot \right ) be the ring of integers modulo 10, and let S be the subset of \mathbb{Z}_{10} represented by {0,2,4,6,8}." My understanding is that \left ( \mathbb{Z}_{10}, +, \cdot \right ) is just notation that they are introducing for this problem, corresponding to the integers up to 10 modulo 10, which is a ring because that's how modulo works. The question goes on to ask about whether the introduced operator applied to S is closed under addition modulo 10. This means that adding any elements of the set and then taking the modulo will yield a member of the set.

Q29 defines a tree as a connected graph with no cycles. I'm actually really confused about what this means. Does it just mean that the graph cannot loop back onto itself? And by extension that differently angled positions of the vertices are considered the same tree?

spablo
Posts: 23
Joined: Thu Oct 27, 2016 10:51 am

Re: GR1268 Q12, 23, 29

Postby spablo » Thu Oct 27, 2016 11:08 am

Q12: This question is asking how many distinct groups of order n are non-isomorphic. There is some theorem you learn in group theory that says that any group of prime order is cyclic. And any two cyclic groups of the same order must be isomorphic since they are generated by one element of order n. So in this case, there is only one group of order 3,5,7, and 11.
You may not have learned the structure theorem for abelian groups. But basically you can determine the number of abelian groups of order n with the rule: if n=mk then gcd(m,k)=1 if, and only if, Z/mZ x Z/kZ ≅ Z/nZ.

For n=4, Z/4Z is an abelian group of order 4, and Z/2Z x Z/2Z is an abelian group of order 4, but since gcd(2,2)=2, then Z/4Z and Z/2Z x Z/2Z are not isomorphic.

That helps you eliminate all but two choices.

For n=9, 9=3*3 and since gcd(3,3)=3, then Z/9Z is not isomorphic to Z/3Z x Z/3Z.

Q23: Yes, that's the right idea. A set S is closed under an operation (+) means that if a and b are in S, then a+b is also in S.

Q29: Yes, a graph with no cycles means you cannot travel on a path and get back to a point you already passed (so yes, no loops). And yes, differently angled positions of the vertices are the same tree (i.e. isomorphic, they have the same structure).




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