1
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ What is the group operation here? $\endgroup$
    – cqfd
    Commented Jul 9, 2019 at 14:48
  • 1
    $\begingroup$ A set only becomes a group after we have defined in it a binary operation for which certain conditions hold. $\endgroup$ Commented Jul 9, 2019 at 14:48
  • $\begingroup$ I am guessing the group is given by multiplication modulo 16... $\endgroup$
    – Asaf Karagila
    Commented Jul 9, 2019 at 16:37

1 Answer 1

1
$\begingroup$

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.

For some further reading on this please see the following:

A very similar example

Is a coset a subgroup?

Canonical coset representatives

$\endgroup$
3
  • $\begingroup$ Thanks for your excellent answer!! Yes, your supplemental information covers everything that I lack. $\endgroup$
    – mallea
    Commented Jul 9, 2019 at 15:32
  • $\begingroup$ By the way, does $\{1, 3, 9, 11 \} \cong \{ 5, 7, 13, 15\}$hold? $\endgroup$
    – mallea
    Commented Jul 9, 2019 at 15:34
  • $\begingroup$ 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. $\endgroup$ Commented Jul 9, 2019 at 15:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .