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?
Thanks in advance!
Best Answer
You want products of the form:
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 $b \in \left<a\right>$.
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.