Help Understanding Why $D_{10} = C_2$ $\times$ $D_5$

Question

I was running around the internet looking at Dihedral groups and ran into a page that claimed that $D_{10} = C_2$ $\times$ $D_5$ (I know notation for Dihedral groups is confusing - here I'm using $D_{n}$ to denote the Dihedral group of order $2n$), and was wondering as to why this is the case.

Now I've been mostly thinking about this in terms of group representations. I know $D_{10} =$ <$a, b: a^{10} = b^2 = 1,$ and $bab^{-1} = a^{9}$>, or equivalently, <$a, b: a^{10} = b^2 = (ba)^2 = 1$> ($a$ is akin to a rotation and $b$ is akin to a reflection).

I also know that $D_5 =$ <$u, v: u^5 = v^2 = 1,$ and $vuv^{-1} = u^4$>, while $C_2 = <c:c^2=1>$. Then thinking of '$\times$' as creating an ordered pair, to me it would seem that $C_2$ $\times$ $D_5$ gives us a group with 3 generators, with the relations listed as above.

I'm not quite sure where to go from here. One thing that stands out to me is the fact that we have two generators of order 2 ($v$ and $c$), and I suspect we might make use of that somehow. However, I'm still not sure how to get the group presentation to look like $D_{10}$'s. I guess thinking about it now, I'm not even really sure how to picture what $C_2$ $\times$ $D_5$ would look like (and if there even is a way to visualize it without $D_{10}$).

Any help would be appreciated, and if there is a different way to approach this problem (Isomorphism between the two groups), that would be welcome as well, though I really do want to try and figure this question out through group presentations as I feel it's a very useful tool.

Answer 1

Think of $D_{10}$ as the symmetry group of the regular decagon. Colour the decagon's vertices alternately red and blue. The red vertices form the vertices of a regular pentagon (the "red pentagon") and similarly with the blue.

Each symmetry in $D_{10}$ either maps the red pentagon to the red and the blue to the blue, or swaps the red and the blue. The $D_5$ is the subgroup of $D_{10}$ taking the red pentagon to itself. It has index two in the $D_{10}$.

Consider the half-turn $t$ in $D_{10}$. It is central in $D_{10}$; it commutes with all operations on $D_{10}$. Moreover, as $5$ is odd, it swaps the red and blue pentagons. The $C_2$ is the group generated by $t$.

As $C_2\cap D_5=\{e\}$, the elements of $C_2$ commute with the elements of $D_5$, and $|D_{10}|=|C_2||D_5|$, then $D_{10}$ is the internal direct product of $C_2$ and $D_5$.

Answer 2

Let's start with a Lemma :

Lemma : $G$ is a group with $H \triangleleft G, K < G$ such that $H \cap K = \left\lbrace e \right\rbrace, HK = G$, then $G \cong H \rtimes_{C_{g}} K$, where $C_{g}$ is the conjugacy.

Let us consider the subgroup $H = \langle r^{5} \rangle$, you can prove that $H \subset Z(D_{10})$, infact there is a more general result about the $Z(D_{n})$

Now let's take $r^{2}$ which has order $5$, (to claim its existence we can use Cauchy Theorem) and consider the subgroup $K$ generated by $r^{2}$ and a reflection $s$, we can see that $r^{2}s = sr^{-2} = s(r^{2})^{-1}$, hence by the lemma since (think about the cardinality of it) the intersection of the subgroups generated by $s^{2},r^{2}$ is trivial, hence $K \cong D_{5}$.

Now let's consider $H$, and apply the Lemma,$K$ is normal since it has index $2$ and $H$ commute with every element in $D_{10}$, including $K$. From the Lemma we have that $D_{10} \cong K \rtimes H \cong D_{5} \times \mathbb{Z}_{2}$, since the semidirect product result to be trivial thanks to $H$ being a subgroup of the center.