Contemporary Abstract Algebra?

Question

What cycle is (a1a2...an)^-1?

Permutation Group

Can someone explain me how to solve this problem, I am a little weak on this subject. Please dont just solve the equation, need details how is done

Answer 1

one way of looking at it is this: the n-cycle (a_0a_1...a_{n-1}) takes a_i to a_{i+1 mod n}, so it's inverse will be the permutation that takes a_i to a_{i-1 mod n}. but this is just the cycle (a_{n-1}a_{n-2}...a_1a_0). i also showed in this answer (http://answers.yahoo.com/question/index;_ylt=AnMUgtquumhWHAaY4d6HJ3Lty6IX;_ylv=3?qid=20071111182353AARKLI6&show=7#profile-info-c5aP5RQnaa)
that (a_1a_2...a_n)^n is the identity for any such n-cycle, so (a_1...a_n)^(-1)=(a_1...a_n)^(n-1) is another way of finding the inverse of an n-cycle.

Answer 2

The inverse of (a1a2..an) is (ana_(n-1) ... a3a2a1)
In other words, just reverse your original cycle.
To see this trace what happens to all the a_i.
a1→a2 in the first cycle and the second sends a2 back
to a1, so a1 is fixed.
Try this for every ai and you will see that this
product leaves every ai fixed.
If you have problems here, look at a particular example:
(123)(321)
Note that 1→2→1
2→3→2
and
3→1→3
so this product leaves all a_i fixed.
Hope that helps!

Answer 3

enable |G|=6n for some n>0 and g be the generator of G. So g^(6n)=a million. assume an element g^ok (ok<6n) in G has order 6, then g^(k6)=a million. yet then ok must be a special of n, enable ok=mn. on the grounds that ok<6n, m<6. apart from, m is comparatively best to 6, in any different case the order of g^ok=g^(mn) isn't 6. to that end, m=a million or 5 and we've precisely 2 factors that have order 6. The case for 8 is completed further.