# Set of representatives of coset in quotient group

Is there any rules to find a set of representative??

I understand a set of representative is a kind of notion for a set that contains many subsets (whose size is larger 1) as elements.

I'm wondering how to find those representatives. And What is the representative for?

Consider the following example:

We have a set (group) $$G$$ $$G = \{1,3,5,7,9,11,13,15\}$$ Now, we have $$\langle 3 \rangle = \{1, 3, 9, 11\}$$.

So we can have the quotient group $$Z = G/\langle 3 \rangle = \{\{1, 3, 9, 11\}, \{5, 7, 13 ,15\}\}.$$

We now have two groups as elements of $$Z$$. What is the representatives of those elements? I think $$3$$ would be the one for the former subgroup. If so, why not the others? $$1, 9, or 11$$. Also what is the representative of the latter group?

• What is the group operation here?
– cqfd
Commented Jul 9, 2019 at 14:48
• A set only becomes a group after we have defined in it a binary operation for which certain conditions hold. Commented Jul 9, 2019 at 14:48
• I am guessing the group is given by multiplication modulo 16... Commented Jul 9, 2019 at 16:37

Your question is not completely clear as you have not specified what the group operation is.

However from looking at your question, I think that your group $$G$$ is the group of units modulo 16. This is sometimes denoted as $$\mathbb{Z}_{16}^{*}$$. The binary operation is multiplication mod 16.

Now assuming this is the group you mean, I will now work through your question.

The group generated by $$\langle 3 \rangle$$ does indeed consist of $$\{1,3,9,11\}$$. As your group $$G$$ is abelian, you know that the subgroup $$\langle 3 \rangle$$ is normal, and so the quotient is well defined.

You have then correctly found the two cosets of $$\langle 3 \rangle$$ in $$G$$ (or put another way, the two elements of the quotient group).

In your question you refer to these as two groups, but this is incorrect, these are two elements of a (quotient) group. Alternatively you might refer to them as cosets (of $$\langle 3 \rangle$$ in $$G$$ ). Notice that in general a coset of $$G$$ is not a subgroup of $$G$$, for instance $${5,7,13,15}$$ does not contain the identity element.

Now I think what your question boils down to is what representatives can you pick. Well you can pick anything that is in the coset as a representative, although sometimes there is a natural choice.

Notice $$\{1,3,9,11\}=1 \cdot \langle 3 \rangle = 3 \cdot \langle 3 \rangle = 9 \cdot \langle 3 \rangle = 11 \cdot \langle 3 \rangle$$ . So here you could pick as your representative any of $$1$$ , $$3$$, $$9$$ or $$11$$.

And $$\{5, 7, 13, 15\}= 5 \cdot \langle 3 \rangle = 7 \cdot \langle 3 \rangle = 13 \cdot \langle 3 \rangle = 15 \cdot \langle 3 \rangle$$ . Here you could pick as your representative any of $$5$$ , $$7$$ ,$$13$$ , $$15$$.

Sometimes a more natural choice will become apparent, for instance if you have a coset containing even numbers and another containing odd numbers, perhaps $$0$$ and $$1$$ would be a natural choice, however you could of course choose something else!

I hope this helps and that I have correctly understood what you are asking.

• By the way, does $\{1, 3, 9, 11 \} \cong \{ 5, 7, 13, 15\}$hold? Commented Jul 9, 2019 at 15:34
• Do you mean by $\cong$ "isomorphic as groups" with the operation we have been using in the answer? If this is what you mean, then the answer is definitely no. Remember the $\{5,7,13,15\}$ is not a group (it does not contain the identity), and so cannot be isomorphic to anything. Let me know if anything is unclear or if I have misunderstood you. Commented Jul 9, 2019 at 15:37