TITLE: Let $n \geq 1$ be an odd integer. Show that $D_{2n}\cong \mathbb{Z}_2 \times D_n$.
QUESTION [14 upvotes]: Let $n \geq 1$ be an odd integer. Show that $D_{2n}\cong \mathbb{Z}_2 \times D_n$. I define a map $$\phi:D_{2n} \rightarrow \mathbb{Z}_2 \times D_n$$ by $R \mapsto (0,r^{\frac{n+1}{2}})$ and $M \mapsto(1,m)$. Then I am stuck at showing the map is bijective. By the way, do we need to show that $(0,r^{\frac{n+1}{2}})(1,m)(0,r^{\frac{n+1}{2}})^{-1}=(1,m)$ ? If we can prove this, then what can we conclude?

REPLY [12 votes]: Let's look at your map--let's call it $\phi$. I claim that it fails to be surjective. Suppose it is surjective, so in particular, $\phi(R^jM^k)=(1,e)$ for some natural $j,k$. Now, $$\phi(R^j)=\phi(R)^j=\bigl(0,r^{\frac{n+1}2}\bigr)^j=\bigl(j\cdot 0,r^{j\cdot\frac{n+1}2}\bigr)=\bigl(0,r^{j\cdot\frac{n+1}2}\bigr)$$ and $$\phi(M^k)=\phi(M)^k=(1,m)^k=(k\cdot 1,m^k)=(k,m^k),$$ so $$\begin{align}(1,e) &= \phi(R^jM^k)\\ &= \phi(R^j)\phi(M^k)\\ &= \bigl(0,r^{j\cdot\frac{n+1}2}\bigr)(k,m^k)\\ &= \bigl(0+k,r^{j\cdot\frac{n+1}2}m^k\bigr)\\ &= \bigl(k,r^{j\cdot\frac{n+1}2}m^k\bigr).\end{align}$$ Then $k$ must be odd (since $1=k$ modulo $2$), but if $k$ is odd, then we can't have $r^{j\cdot\frac{n+1}2}m^k=e$, so we have a contradiction.

Finding an isomorphism is (in many cases) quite a bit out of our way, but let's go ahead and do it here anyway. Whatever we map $R$ to is going to have to have order $2n$ for the map to be injective. (Why?) A convenient choice is $(1,r),$ because $$(0,e)=(1,r)^k=(k,r^k)$$ implies that $k$ is even (since $k=0$ modulo $2$) and that $n$ divides $k$ (since $r^k=e$ and $r$ has order $n$), whence $k$ is an integer multiple of $2n$ (since $n$ is odd). Whatever we map $M$ to can't be in the subgroup of $\Bbb Z_2\times D_n$ generated by $(1,r)$ (why not?). An easy choice then is $(1,m).$ Now, let's prove surjectivity by showing that every member of a generating set of $\Bbb Z_2\times D_n$ is mapped to by the homomorphism $\Phi:D_{2n}\to\Bbb Z_2\times D_n$ induced by $R\mapsto(1,r),M\mapsto(1,m).$ (Why is this enough?) A nice generating set of $\Bbb Z_2\times D_n$ is $$\bigl\{(1,e),(0,r),(0,m)\bigr\},$$ and it can be determined that $$\Phi(R^n)=(1,e),\;\Phi(R^{n+1})=(0,r),\text{ and }\Phi(R^nM)=(0,m).$$ Since we're dealing with finite groups of the same order, then surjectivity is the same as injectivity, so we're done.

The approach above was fairly tedious, and we'd generally like to avoid using it. Instead, consider the following
Theorem: Let $G,H,K$ be (multiplicative) groups. Then $G\cong H\times K$ if and only if $G$ has normal subgroups $H',K'$ such that $H'\cong H,K'\cong K,H'\cap K'=\{\text{id}_G\},$ and $$G=H'K':=\{hk:h\in H',k\in K'\}.$$
Proving the above result is a nice exercise (let me know if you want any hints on how to prove the backward direction). It then yields the following
Corollary: Let $G,H,K$ be finite (multiplicative) groups. Then $G\cong H\times K$ if and only if $G$ has normal subgroups $H',K'$ such that $H'\cong H,K'\cong K,\text{ and }H'\cap K'=\{\text{id}_G\}.$
By the Corollary, we need only find normal subgroups of $D_{2n}$ isomorphic to $\Bbb Z_2$ and $D_n$, whose intersection is precisely $\{E\}.$ The subgroup generated by $R^2$ and $M$ should do the trick as a subgroup isomorphic to $D_n$ (automatically normal, since it has index $2$), and since $R^n$ has order $2$ and commutes with everything, then the subgroup it generates is normal and isomorphic to $\Bbb Z_2$.

Edit: Sadly, the Corollary is false in the form shown above! (Credit to Geralt of Rivia for pointing out the issue.) Another condition is needed. If we add the hypothesis that $\|H\|\cdot\|K\|=\|G\|,$ that gets us the rest of the way.
As far as the specific example goes, since we readily have $\left\|D_{2n}\right\|=2\cdot\left\|D_n\right\|,$ then the adjusted Corollary gets us there.