# Finding Left and Right Cosets

I'm having a little bit of trouble calculating cosets. I understand that the intuitive definition of cosets are the "slices of bread if the bread is a group," and the formal defintions are:

gH = { gh : h an element of H }, Hg = { hg : h an element of H }

But how do I actually calculate the cosets? For instance,

Consider K = {$R_0$, $R_{180}$} $\le$ $D_4$, where $D_4$ is the group of the symmetries of a square, and $R_0$ and $R_{180}$ are rotations by 0 and 180 degrees clockwise, respectfully. What are the left and right cosets of K?

Do I just simply multiply each element of $D_4$ by each element of K, on left and right sides? Any help would be great. Thank you.

• Where did you find the useless analogy that cosets are "slices of bread if the bread is a group"? No wonder you are confused: that description doesn't tell you anything. – symplectomorphic Oct 14 '16 at 4:31
• @symplectomorphic Presumably you have a better analogy? – PeterJL Oct 14 '16 at 4:59

what you should do is that take an element from $D_4$, say $g$, and $gK=\{gk:\forall k\in K\}$, while $Kg=\{kg:\forall k\in K\}$. I can see that you understand that.

As an example, let $g=R_{90}$, according to your denotation. $gK=\{R_{90}R_0,R_{90}R_{180}\}=\{R_{90},R_{270}\}$. $Kg=\{R_0R_{90},R_{180}R_{90}\}=\{R_{90},R_{270}\}$ , and you can do that for every element.

Actually, having an instinct of slicing bread is not a bad idea. Actually you slice the entire group $D_4$ into equivalent classes, when you try to create cosets. This process is called partition. The equivalent relation is defined as $a\sim b$ when $a$ and $b$ belongs to the same coset. You can even define a group structure on the set of cosets if $K$ is a normal subgroup, which is right in your example.

Hope that helped.

• So H is the group K, and we take each element of $D_4$ and multiply it to the left and right of K? – Max Oct 14 '16 at 5:25
• H is a given subgroup (or more generally subset) of a given group. In this case, H is {R_0, R_{180}} while the group is $D_4$. Well in this problem I interpreted your subgroup K as H, and I used a wrong denotation. Sorry for the confusion. – Junkai Dong Oct 14 '16 at 5:29
• @Max I fixed the denotation. Is it clearer to you this time? – Junkai Dong Oct 14 '16 at 5:30
• Ah yes, it's very helpful. Thank you! – Max Oct 14 '16 at 5:31
• @Max I am glad that I helped. – Junkai Dong Oct 14 '16 at 5:31