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
GR 9367 Question 22

 Posts: 47
 Joined: Mon Mar 22, 2010 2:42 am
Re: GR 9367 Question 22
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.
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.
Re: GR 9367 Question 22
speedychaos4 is right. In case you would like another explanation, here is one.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
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 piecebypiece, in reverse order, so that everything simplifies to the identity.

 Posts: 27
 Joined: Tue Apr 06, 2010 8:22 am
Re: GR 9367 Question 22
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.
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.