Factor group order

• Nov 7th 2009, 11:58 AM
sfspitfire23
Factor group order
If $K\lhd G$ and $|g|=n$, $g\in G$, show that the order of $Kg$ in $G/K$ divides $n$.

So I know that $G=\{1,g,\cdots g^n\}$ and will $K=\{1,g, \cdots g^{n-1}\}$ ? From Lagrange's theorem K|G?

Thanks guys
• Nov 7th 2009, 12:10 PM
tonio
Quote:

Originally Posted by sfspitfire23
If $K\lhd G$ and $|g|=n$, $g\in G$, show that the order of $Kg$ in $G/K$ divides $n$.

So I know that $G=\{1,g,\cdots g^n\}$ and will $K=\{1,g, \cdots g^{n-1}\}$ ? From Lagrange's theorem K|G?

Thanks guys

Two easy lemmas for you to remember/prove:

1) In a finite group the order of any element divides the order of the group;

2) If $f: G\rightarrow H$ is a group homomorphism, then for any $g\in G\,,\,\,ord(f(g))\mid ord(g)$ , and if $ord(g)=\infty \,\,then\,\, ord(f(g))=\infty \,\,or\,\,else\,\,f(g)=1$

Tonio
• Nov 7th 2009, 02:45 PM
alunw
Let G be the group of integers under addition and H the integers modulo n under addition. That is surely a homomorphism that is sending torsion free elements to elements with finite order that are not the identity. Am I misunderstanding your notation?
• Nov 7th 2009, 03:35 PM
tonio
Quote:

Originally Posted by alunw
Nop, you're right and I got confused. It should be that if $ord(g)=\infty$ then all the options are open.
For any element g in G $Kg=\{kg|k\in K\}$ defines the right coset of K. Each k in K will have a different product when multiplied by any g. Thus, each element of K will create a corresponding unique element of Kg. So, Kg will have the same number of elements as K (same order as K). Now, the order of K will be $k^m=1, |k|=m$ and the order of G is $g^m=1$. They each produce the identity element and thus divide each other.....?