How to find the left and right cosets of the subgroup $H = \{r_0, s_0\}$ of $D_4$? And are they the same?
If we let H' = \{r_0,r_2\}, are the left and right cosets the same, where H' is a subgroup of $D_4$?
Thanks
How to find the left and right cosets of the subgroup $H = \{r_0, s_0\}$ of $D_4$? And are they the same?
If we let H' = \{r_0,r_2\}, are the left and right cosets the same, where H' is a subgroup of $D_4$?
Thanks
By actual computation using the multiplication table at the link that you gave, we have $\begin{align*} &Hr_0=\{r_0,s_0\}\text{ and }r_0H=\{r_0,s_0\}\\ &Hr_1=\{r_1,s_3\}\text{ and }r_1H=\{r_1,s_1\}\\ &Hr_2=\{r_2,s_2\}\text{ and }r_2H=\{r_2,s_2\}\\ &Hr_3=\{r_3,s_1\}\text{ and }r_3H=\{r_3,s_3\}\\ &Hs_0=\{s_0,r_0\}\text{ and }s_0H=\{s_0,r_0\}\\ &Hs_1=\{s_1,r_3\}\text{ and }s_1H=\{s_1,r_1\}\\ &Hs_2=\{s_2,r_2\}\text{ and }s_2H=\{s_2,r_2\}\\ &Hs_3=\{s_3,r_1\}\text{ and }s_3H=\{s_3,r_3\}\;. \end{align*}$
If you go carefully through both columns, you’ll see that each left coset and each right coset appears twice. The right cosets are
$\begin{align*} &Hr_0=\{r_0,s_0\}=\{s_0,r_0\}=Hs_0\\ &Hr_1=\{r_1,s_3\}=\{s_3,r_1\}=Hs_3\\ &Hr_2=\{r_2,s_2\}=\{s_2,r_2\}=Hs_2\\ &Hr_3=\{r_3,s_1\}=\{s_1,r_3\}=Hs_1\;, \end{align*}$
and the left cosets are
$\begin{align*} &r_0H=\{r_0,s_0\}=\{s_0,r_0\}=s_0H\\ &r_1H=\{r_1,s_1\}=\{s_1,r_1\}=s_1H\\ &r_2H=\{r_2,s_2\}=\{s_2,r_2\}=s_2H\\ &r_3H=\{r_3,s_3\}=\{s_3,r_3\}=s_3H\;. \end{align*}$
As you can see, the left and right cosets of $H$ are not all the same: $Hr_1=Hs_3$ and $Hr_3=Hs_1$ are right cosets that are not left cosets, and $r_1H=s_1H$ and $r_3H=s_3H$ are left cosets that are not right cosets.
I’ll leave H' for you to try on your own, at least for now.