1
$\begingroup$

I am trying to understand the concept of factor group. The definition of factor group I know is the following: Let $G$ be a group and $H$ be a subgroup of $G$. Then the group of cosets denoted by $G/H$ is called the factor group of $G$ by $H$. Now I am looking at an example: Let $G=\mathbb{Z}$ and $H=5\mathbb{Z}$. Then I find the cosets of $5\mathbb{Z}$. Here they are:

  • Cosets of $5\mathbb{Z}$ containing $0$, $\{\ldots,-10,-5,0,5,10,\ldots\}$
  • Cosets of $5\mathbb{Z}$ containing $1$, $\{\ldots,-9,-4,1,6,11,\ldots\}$
  • Cosets of $5\mathbb{Z}$ containing $2$, $\{\ldots,-8,-3,2,7,12,\ldots\}$
  • Cosets of $5\mathbb{Z}$ containing $3$, $\{\ldots,-7,-2,3,8,13,\ldots\}$
  • Cosets of $5\mathbb{Z}$ containing $4$, $\{\ldots,-6,-1,4,9,14,\ldots\}$

My question is, is the factor group $\mathbb{Z}/5\mathbb{Z}$ the union of all these $5$ cosets? Then I think $G/5\mathbb{Z} =\mathbb{Z}$ since the union of these cosets are equal to $\mathbb{Z}$. Can anyone help me with this?

Thanks.

$\endgroup$
5
$\begingroup$

The factor group is not the union of cosets, it's the set of cosets, so $${\bf Z}/5{\bf Z}=\{5{\bf Z},1+5{\bf Z},2+5{\bf Z},3+5{\bf Z},4+5{\bf Z}\}$$

The union of cosets is the original group. And $H$ has to be a normal subroup for the set of cosets to be a group (this is trivial if $G$ is abelian, of course).

$\endgroup$
  • $\begingroup$ and it is a funny coincidence if you are Tomasz from RSI 2007, he was also from poland $\endgroup$ – Yasin Razlık May 25 '13 at 14:48
  • $\begingroup$ @bigO: I don't even know what RSI 2007 is, so I'm likely not the same person. ;) $\endgroup$ – tomasz May 25 '13 at 15:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.