# Dihedral group $D_3$(also $D_4$) - reflection compose rotation

I’ve read in a book(image link)that for dihedral group $$D_3, a \circ r_1 = b$$where a means reflection about $$AO$$ ($$O$$ is the centroid of an equilateral triangle $$ABC$$), $$r_1$$ means rotation about O through $$120^\circ$$. Shouldn’t this binary composition turn out to be $$c$$? I’ve got c both by physically performing the operations (reflection followed by rotation on an equilateral triangle shaped piece of paper) and also by multiplying the permutations $$a$$ and $$r_1$$, i.e., $$ar_1$$. In the image which I’ve given, my confusion lies in the entire marked part. Same kind of confusion I’m having with $$D_4$$. Please help!

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A Ghosh is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

## 3 Answers

Yours is a common confusion. When you physically move the triangle you have to choose a convention about the labels. Either they move with the triangle or they are fixed and the triangle moves over them. Whichever way you are doing it the book is doing the other.

• But even after following the book’s way I’m getting it reversed. Shouldn’t reflection about AO followed by rotation through 120deg (anti-clockwise) be same as reflection about CO? Commented Jun 11 at 6:53

I get $$b$$.

Label the corners $$1,2,3$$ going anticlockwise from $$A$$. Then $$a$$ is the permutation $$(23)$$ and $$r_1$$ is, assuming anticlockwise rotation, $$(123)$$.

Now $$a\circ r_1\mapsto (23)\circ(123)=(13)\mapsto b.$$

This exploits an isomorphism, $$D_3\cong S_3$$.

consider this image .You are taking rotation anti clockwise, but they are taking it clockwise, I think that's what making the confusion.

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mrinal nath is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.