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Let $G = \mathbb Z_3 \times \mathbb Z_6$, $H = \langle (1,2)\rangle$ and let $K = \langle (1,3)\rangle$. List all cosets of $H$ and $K$.

Can somebody please explain me how to do this problem. First of all I'm having trouble how to obtain the set of $H$ and $K$.

Definition: Let H be a subgroup of the group $G$, and let a element of $G$. The set

$aH$ = {x element of G | x = ah for some h element of H}

is called the left coset of $H$ in $G$ determined by $a$. Similarly, the right coset of $H$ in $G$ determined by $a$ is the set

$Ha$ = {x element of G |x = ha for some h element of H}.

The number of left cosets of $H$ in $G$ is called the index of $H$ in $G$, and is denoted by $[G : H]$

I need some explanation on how to do this kind of problems I really want to understand. Any help will be appreciated.

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  • $\begingroup$ The first step is figuring out what $H$ and $K$ are. Do you know what the notation $\langle g\rangle$ means? Use your understanding of that notation to start listing out elements in $H$ and $K$ respectively, until you've got them all. The next step is to compute the index of $H$ and $K$ in $G$ (we do this to see ahead of time how many cosets to look for). There is a formula for $[G:H]$ in terms of $|G|$ and $|H|$, do you know it? Finally you want the cosets. To do this, test for candidates for transversals in $\langle(1,0)\rangle$ or $\langle(0,1)\rangle$ for example, or something else clever. $\endgroup$
    – anon
    Oct 13 '13 at 4:00
  • $\begingroup$ Of what I understand it means that $<g>$ is a generator of G, right? $\endgroup$ Oct 13 '13 at 4:05
  • $\begingroup$ No. It means $\langle g\rangle$ is generated by $g$. $\endgroup$
    – anon
    Oct 13 '13 at 4:06
  • $\begingroup$ the set $<g>$ is the cyclic subgroup generated by $g$. $\endgroup$ Oct 13 '13 at 4:11
  • $\begingroup$ Yes. So $\langle g\rangle=\{g^{-2},g^{-1},g^0,g^{+1},g^{+2},\cdots\}$ (the negative powers many be omitted if the group is finite, or less restrictively, if $g$ has finite order). So $\langle g\rangle=\{e,g,g^2,\cdots\}$ here. $\endgroup$
    – anon
    Oct 13 '13 at 4:17
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Let's do this for $H$, and leave $K$ as a similar exercise :

  1. $H = \langle (1,2)\rangle = \{(1,2), (2,4), (0,0)\}$ since $$ (3,6) \equiv (0,0) $$ You need to check that there are no other elements. Hence, $|H| = 3$ and $[G:H] = 18/3 = 6$. Thus, there are 6 cosets you need to find.

  2. The cosets are all of the form $(i,j) + H$, where $(i,j) \in G$. $$ (0,0) + H, (1,0) + H, \ldots, (2,4)+H, (2,5) + H $$ There are 18 such elements; but there are overlaps : $$ (i,j) + H = (k,l) + H \Leftrightarrow (i-k,j-l) \in H $$ $$ \Leftrightarrow (i-k,j-l) \in \{(1,2), (2,4), (0,0)\} $$

  3. Now we list the cosets : $$ (0,0) + H = (1,2) + H = (2,4) + H $$ $$ (1,0) + H = (2,2) + H = (3,4) + H $$ $$ (2,0) + H = (0,2) + H = (1,4) + H $$ $$ (0,1) + H = (1,3) + H = (2,5) + H $$ $$ (1,1) + H = (2,3) + H = (3,5) + H $$ $$ (2,1) + H = (0,3) + H = (1,5) + H $$ These are all the cosets of $H$. Can you do the same thing with $K$?

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  • $\begingroup$ Thank you so much. Now that you explained I do. $\endgroup$ Oct 14 '13 at 6:06

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