# Unique factorization of dihedral group

My goal is to prove the following about the dihedral group $$D_{2n}$$:

Prove that every element in $$D_{2n}$$ has a unique factorization of the form $$a^{i}b^{j}$$, where $$0 \leq i < n$$ and $$j=0$$ or $$1.$$

I know that the cyclic subgroup $$\left \langle a \right \rangle$$ has order $$n.$$ From this, I know that this has index $$2.$$ Thus $$D_{2n}$$ is the disjoint union $$\left \langle a \right \rangle \cup \left \langle a \right \rangle b.$$

After this, I am pretty stuck. Am I headed in the right direction? What would be the correct way to finish this proof?

• If you already know that $D_{2n}$ is a disjoint union of $\langle a\rangle$ and $\langle a\rangle b$, then why don't you already know that every element can be written as $a^ib^j$ with $j=0$ or $1$, $0\leq i\lt n$? And once you know there is at least one such factorization, what is the difficulty in proving uniqueness? – Arturo Magidin Mar 7 at 22:44
• So, I gather from what you are saying that the proof is almost complete. I just need to show it is unique? – MathIsLife12 Mar 7 at 22:57
• Well, you also need to realize why you've already proven the existence of the factorization... – Arturo Magidin Mar 7 at 22:58

$$a^{i}b^{j}$$, where $$0 \leq i < n$$ and $$j=0$$ or $$1.$$
there are $$2n$$ of these: $$a^0,a^1,\ldots,a^{n-1}$$ and $$a^0b,a^1b,\ldots,a^{n-1}b$$. These $$2n$$ products each give you an element of $$D_{2n}$$ so as long as they all represent distinct elements, the factorization is unique.
The $$n$$ products of the form $$a^i$$ are distinct because $$|a| = n$$. Likewise, the elements $$a^ib$$ are distinct because $$a^{i_1}b = a^{i_2}b$$ implies $$a^{i_1} = a^{i_2}$$ by cancellation, and then $$i_1 = i_2$$ since $$0\leq i \leq n$$. Finally, there's no $$a^{i_1} b = a^{i_2}$$ because that would imply that $$b \in \left$$.
That means your $$2n$$ products are in 1-to-1 correspondence with the elements of $$D_{2n}$$; each element of $$D_{2n}$$ has a unique representative as one of the products.