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I'm trying to understand how cosets work. I came across with the following question:

Check if there is a subgroup of order $2$ of $U_{20}$ (Euler group) and find its cosets.

In the solution they found that $U_{20}$ has three subgroups of order $2$: $\langle 9\rangle$, $\langle 11\rangle$ and $\langle 19\rangle$. Later they took the group $\langle 9\rangle$ and try to find its cosets in $U_{20}$ - they just said that the cosets are: $\langle 9\rangle, \{3,7\},\{11,19\},\{13,17\}$ without explaining why. I'm familiar with the theorem that the left cosets are $gH=\{gh\,:\,h\in H\}$ and right cosets are $Hg=\{hg\,:\,h\in H\}$, but I don't understand how this gives me the solution. I also know that from Lagrange we get:

$$ [U_{20}\,:\,\langle 9\rangle]=\frac{|U_{20}|}{|\langle 9\rangle|}=\frac{8}{2}=4$$

My question is how they understood that the cosets are $\langle 9\rangle, \{3,7\},\{11,19\},\{13,17\}$?

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  • $\begingroup$ $U_{20}=\{1,3,7,9,11,13,17,19\}.$ The first coset is always your subgroup $H=\{1,9\}.$ Now, any new coset will be disjoint, so we need a coset with $3$ in it. Take $3H=\{3,27\}=\{3,7\}$ will be a coset with $3$ in it. We've already found a coset with $1,3,7,9$ in it, so try $11H=\{11,99\}=\{11,19\}.$ Finally, the two remaining are $\{13,17\}=13H.$ $\endgroup$ – Thomas Andrews Jul 9 '19 at 19:48
  • $\begingroup$ when multiplication is commutative (as for invertible elements modulo $20$), left and right cosets are the same $\endgroup$ – J. W. Tanner Jul 9 '19 at 19:53
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That is not a theorem but the definition of a coset. They probably did not explain it because they just computed the cosets. For example for $g = 3$ we get $$g \cdot \langle 9 \rangle = 3 \cdot \langle 9 \rangle = 3 \cdot \lbrace 9, 1 \rbrace = \lbrace 3 \cdot 9, 3 \cdot 1 \rbrace = \lbrace 7,3 \rbrace.$$ Now just do the computation for the other elements (not being in the cosets so far) and you will see the four cosets.

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    $\begingroup$ all cosets have the same size always? $\endgroup$ – vesii Jul 9 '19 at 19:52
  • $\begingroup$ @vesii See math.stackexchange.com/questions/187572/… $\endgroup$ – ArsenBerk Jul 9 '19 at 19:54
  • $\begingroup$ Yes. That is because you can define a bijection in the following way. Let $aH$ and $bH$ be cosets. Then consider the map $f \colon aH \rightarrow bH$, $ah \mapsto bh$ (the multiplication with $ba^{-1}$ from the left). This map has an inverse given by the multiplication with $ab^{-1}$. $\endgroup$ – Con Jul 9 '19 at 19:55
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Note that $U_{20} = \{1,3,7,9,11,13,17,19\}$ and define $H = \langle 9\rangle = \{1,9\}$. Now, as definitions of cosets you gave, we have (note that all calculations are done in modulo $20$) $$1H = H,\ 3H = \{3\cdot1,3\cdot9\} = \{3,7\},\ 11H = \{11\cdot1,11\cdot9\} = \{11,19\},\ 13H = \{13,17\}$$

Note that for instance once we find $3H$ to be $\{3,7\}$, we don't check $7H$ again because they must be the same (since $7 \in 3H$). So these four checks are enough.

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One coset will be the subgroup itself. Now take an element of the group that is not in any coset you have so far, for example $3$. Multiply this element with the elements in the subgroup (your group is abelian so you need not worry about left and right cosets here) you will get $\{3,7\}$. Repeat this process, say take $11$. Keep doing this until you have no elements left. If you need more guidance let me know.

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