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Cyclic groups

 
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Feb28-09, 07:38 PM   #1
 

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!
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Mar1-09, 02:35 PM   #2
 
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Quote by hsong9 View Post
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 [itex]g^{20}=e[/itex] then [itex]g^{40}=e[/itex] also.

B. a)H=<a^d> , right?
Yes.

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?
Mar1-09, 02:47 PM   #3
 
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The least common multiple of 20 and 8 is 2*4*5= 40. [itex](g^8)^5= (g^20)^2= e[/itex].

The least common multiple of 3 and 20 is 60. [itex](g^3)^20= (g^20)^3= e[/itex].
Mar1-09, 03:38 PM   #4
 

Cyclic groups


so.. for b) is H=(1 2) * (2 3) = (1 2 3)..?
Mar1-09, 04:10 PM   #5
 
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Yes, that's right.
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