# Understanding how to find the cosets of a group

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 $$\langle 9\rangle, \{3,7\},\{11,19\},\{13,17\}$$?

• $U_{20}=\{1,3,7,9,11,13,17,19\}.$ 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 $\{13,17\}=13H.$ – Thomas Andrews Jul 9 '19 at 19:48
• when multiplication is commutative (as for invertible elements modulo $20$), left and right cosets are the same – J. W. Tanner Jul 9 '19 at 19:53

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.
• 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 $ab^{-1}$. – Con Jul 9 '19 at 19:55
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.
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.