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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

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$D_4$ is a small group, and $H$ and $H'$ are very small; have you tried simply writing out the cosets? The right coset of $H$ by $r_2$, for instance, is $Hr_2=\{r_0r_2,s_0r_2\}$; $r_0r_2=r_2$, and $s_0r_2=$ what? Now do this for each of the $8$ elements of $D_4$, and do the same thing on the lefthand side, and compare the two collections of cosets. –  Brian M. Scott Feb 23 '12 at 8:22
    
Is s0r2 = s2, but why? –  James R. Feb 23 '12 at 8:27
    
The left cosets of H by r2, is r2H = {r2r0,r2s0} and r2r0 = r2 and r2s0 = s2. –  James R. Feb 23 '12 at 8:28
    
I can’t be sure, because it depends on how you’re labelling the elements of the group, but it seems likely. You need to write out the group multiplication table, so that you can simply look up each possible product; it’s only $8\times 8$. –  Brian M. Scott Feb 23 '12 at 8:33
    
Can you help out with the element s0 of H, for D4? –  James R. Feb 23 '12 at 8:44

1 Answer 1

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.

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I'm stuck with H' and I'm not seeing what is being multiplied, like for example, s2H = ? –  James R. Feb 23 '12 at 10:58
    
never mind, now getting it! thanks! –  James R. Feb 23 '12 at 11:10

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