# Compute cosets in the dihedral group Stuck on how to answer this. So I understand the dihedral group $D_8$ consists of $16$ elements with $n$ rotations and $n$ reflections. Each of the reflections have order $2$ so $b^2 = 1$ as shown. $a^8 = 1$ since $8$ is the order of the cyclic subgroup of rotations. And the reflection defined by $ba = a^{-1}b$.

Now I get how to compute left cosets ($aH = \{ah\ |\ h \in H\}$), but not sure how to do this for an arbitrary dihedral group.

What exactly is $<a^3>$, the cyclic subgroup of $a^3$ and how would I go about computing that? Counterclockwise rotations by $135$? $\{a^3, a^6, a\}$?

And furthermore, how would I compute the left coset of each element. For example, $a^3 \cdot a^6 =\ ...a^9 = a?$

Would really appreciate clarification as I think I'm misunderstanding something about the dihedral groups, seeing as how I'm unable to compute this.

• Whats the book ? – Aryadeva Oct 24 '17 at 1:53

## 1 Answer

$<a^3>$ is $\{a^3, a^6, a, a^4, a^7, a^2, a^5, 1\} = C_8$. So clearly, $a^6$ and $a^4$ belong to the same coset. Same with the second problem.

$<a^3b> = \{a^3b, 1\}$ because $a^3ba^3b = a^3a^{-1}ba^2b = a^2ba^2b = abab = b^2 = 1$. The cosets are $\left\{\{a^3b, 1\}, \{a^4b, a\}, \{a^5b, a^2\}, \{a^6b, a^3\}, \{a^7b, a^4\}, \{b, a^5\}, \{ab, a^6\}, \{a^2b, a^7\}\right\}$.

• For the second one, not sure how to compute cosets... $a^6b \cdot a^3 = a^6ba^3 = a^6a^{-3}b = a^3b =\ ??$ – SS' Oct 24 '17 at 1:48