# quotient group inheritance of some properties from original group

Let us say that there is the group with addition modulo 6:

$G = \{0,1,2,3,4,5\}$ and let $N = \{0,3 \}$.

Then the quotient group $G/N$ would be $G/N = \{ \{0,3\}, \{1,4 \}, \{2,5\} \}$.

According to my sources, they say that as $N$ is normal subgroup, $G/N$ must be also group. As the group $G$ is abelian, according to my sources (Wikipedia and textbook), they say that $G/N$ must be abelian group.

The question is, by quotient group $G/N$ being group, is group operation done by assuming that in the example, $0$ and $3$ are treated same, $1$ and $4$ are treated same and $2$ and $5$ are treated same? And abelian-ness can be checked by adding $(3+4) \pmod 6 = (4+3) \pmod 6 = 1 = 4?$

Or is this something else?

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Yeah, essentially you are right that you have to treat them the same. So, for example, using the fact that $1$ and $4$ are the same we have $1+2=3\text{ mod }6$ while $4+2=0\text{ mod }6$ and this makes sense as $0$ and $3$ are treated the same. You should realise though that what is really going on is that you are working $\text{mod }3$. –  user1729 Mar 28 '13 at 10:14
You may find interesting this question about quotient sets. –  A.P. Mar 28 '13 at 12:48

## 1 Answer

This is something else: in the quotient you have equivalence classes which you will multiply (sum) according to the rules you were surely told. There are no more mere elements as in the mother group, yet it is always possible to pick up some representative from each class andwork with it...

This subject is better understood in calss, with a decent blackboard and some chalk, and doing many examples, but in your case we can say, for example, that

$$\{1,4\}+\{2,5\}=\{0,3\}\iff \bar 1+\bar 2=\bar 0$$

where the bar above means that we'reworking the equiv. class, not merely with the bare elements.

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The point here is that elements which are in the same equivalence classes are "treated the same". Equivalence classes just formalise this. –  user1729 Mar 28 '13 at 10:17
Yes, but the operation is not more modulo $\,6\,$ , as my example shows, and this is what the OP asked, among other things. –  DonAntonio Mar 28 '13 at 12:47