Physics Forums (http://www.physicsforums.com/index.php)
-   Calculus & Beyond Homework (http://www.physicsforums.com/forumdisplay.php?f=156)

 hsong9 Feb28-09 07:38 PM

Cyclic groups

1. The problem statement, all variables and given/known data
A. Let |g| = 20 in a group G. Compute
|g^2|, |g^8|,|g^5|, |g^3|

B. In each case find the subgroup H = <x,y> of G.
a) G = <a> is cyclic, x = a^m, y = a^k, gcd(m,k)=d
b) G=S_3, x=(1 2), y=(2 3)
c) G = <a> * <b>, |a| = 4, |b| = 6, x = (a^2, b), y = (a,b^3)

3. The attempt at a solution
A. I know |g^2| = 20/2 = 10 and |g^5| = 20/5 = 4
But |g^8|, |g^3| don't know..

B. a)H=<a^d> , right?
but
I don't know how to solve b) and c)
Thanks!

 Tom Mattson Mar1-09 02:35 PM

Re: Cyclic groups

Quote:
 Quote by hsong9 (Post 2096834) 3. The attempt at a solution A. I know |g^2| = 20/2 = 10 and |g^5| = 20/5 = 4 But |g^8|, |g^3| don't know..
Don't forget that if $g^{20}=e$ then $g^{40}=e$ also.

Quote:
 B. a)H= , right?
Yes.

Quote:
 but I don't know how to solve b) and c) Thanks!
b should be easy, because you've got a concrete group to play with. Just get in there and start computing. As for c, what does <a>*<b> mean?

 HallsofIvy Mar1-09 02:47 PM

Re: Cyclic groups

The least common multiple of 20 and 8 is 2*4*5= 40. $(g^8)^5= (g^20)^2= e$.

The least common multiple of 3 and 20 is 60. $(g^3)^20= (g^20)^3= e$.

 hsong9 Mar1-09 03:38 PM

Re: Cyclic groups

so.. for b) is H=(1 2) * (2 3) = (1 2 3)..?

 Tom Mattson Mar1-09 04:10 PM

Re: Cyclic groups

Yes, that's right.

 All times are GMT -5. The time now is 10:06 PM.