GR 9367 Question 22

Forum for the GRE subject test in mathematics.
mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

GR 9367 Question 22

Postby mathQ » Mon Apr 05, 2010 11:03 am

Ques: If b and c are elements in a Group G, and id b^5 = C^3 = e ( e= identity element of G), then the inverse of b^2cb^4c^2 is

A) b^3c^2bc
B) b^4c^2b^2c
C) c^2b^4cb^2
D) cb^2c^2b^4
E) cbc^2b^3


Can anyone please explain the answer?

Thanks

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

Re: GR 9367 Question 22

Postby speedychaos4 » Tue Apr 06, 2010 1:18 am

Hi mathQ,
I'm new to group theory (just glanced at some pages covering basic concepts), but I do have a solution for this question, but I think it is only suitable for this kind of multiple choices questions.

since we have b^5 = C^3 = e from the question, and we are trying to find the inverse of b^2cb^4c^2, we should use the eq b^5 = C^3 = e intuitively.
Let A denote b^2cb^4c^2 and now we determine A^-1 by the eq b^5 = C^3 = e.
If AA^-1=e, we can try writing down A^-1 by the manner that the product of every adjacent factor between A and A^-1 is e, then it is easily to find cbc^2b^3, suppose A^-1=cbc^2b^3, we evaluate A^-1A, then we also have e, therefore, cbc^2b^3 is the inverse that we are looking for.

I'm sorry but it seems that I cannot express what I am thinking clearly due to my bad English.

ArcTan
Posts: 3
Joined: Sat Feb 27, 2010 10:43 am

Re: GR 9367 Question 22

Postby ArcTan » Thu Apr 08, 2010 1:44 am

mathQ wrote:Ques: If b and c are elements in a Group G, and id b^5 = C^3 = e ( e= identity element of G), then the inverse of b^2cb^4c^2 is

A) b^3c^2bc
B) b^4c^2b^2c
C) c^2b^4cb^2
D) cb^2c^2b^4
E) cbc^2b^3


Can anyone please explain the answer?

Thanks


speedychaos4 is right. In case you would like another explanation, here is one.

Since b^5=e, we know the following...
b(b^4)=e
(b^2)(b^3)=e
(b^3)(b^2)=e
(b^4)b=e

And since c^3=e, we also know the following...
c(c^2)=e
(c^2)c=e

Now it is clear what is the inverse of each element.
We construct the inverse of b^2cb^4c^2 piece-by-piece, in reverse order, so that everything simplifies to the identity.

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

Re: GR 9367 Question 22

Postby EugeneKudashev » Thu Apr 08, 2010 1:57 am

I think you might want to keep in mind the following fact: (abcd)^(-1) = d^(-1) c^(-1) b^(-1) a^(-1) [could be generalized for any finite number of terms].
using this thing plus establishing the connections like c^(-2) = c (due to c^3=e) et cetera already mentioned above, it is easy to get the result.




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