In an proof that I recently read, the following 'fact' is used, where $D_{2n}$ denotes the dihedral group of order $2n$:
If $n$ is even, then $D_{2n} \cong C_2 \times D_n$.
The (short) given justification is that the centre $Z(D_{2n}) \cong C_2$, whenever $n$ is even, and is trivial provided that $n$ is odd.
However, here is a result that a friend of mine found in the litterature which contradicts the previous one.
Suppose that 4 divides $n$. Then $D_{2n}$ is not isomorphic to $C_2 \times D_n$.
Proof: Suppose otherwise. We know that $Z(D_{2n}) \cong C_2$. By assumption, $D_{2n} \cong C_2 \times D_n$, therefore $Z(D_{2n}) \cong Z(C_2 \times D_{n})$. Since in a direct product $A \times B$, the groups $A$ and $B$ commute, we obtain that $Z(C_2 \times D_{n}) = Z(C_2) \times Z(D_n) \cong C_2 \times C_2$, a contradiction. QED.
Now comes my first question : Is the above proof correct?
Second question : Given a finite group $G$ whose centre is not trivial, does it always exist a group $H$ such that $G \cong Z(G) \times H$?
If so, then given $n$ a multiple of 4, the dihedral group $D_{2n}$ could be written as a direct product $C_2 \times H$. Here comes my third question : What should be $H$ since we know that it cannot have nontrivial centre (and, in particular, since $H$ cannot be $D_n$)?
Best Answer
I like your questions.
Yes, and it is very silly.
No, for example the quaternion group of order 8 is not like this.
(So, there is no such H).
To be clear, the dihedral group of order 8k+4 is isomorphic to the direct product of a cyclic group of order 2 with a dihedral group of order 4k+2, but all other dihedral groups are "directly indecomposable", that is, they do not have a decomposition as a direct product of two non-identity groups.
In case you want to prove the stronger indecomposability: Basically, you look at the normal subgroups. Assuming the order of the dihedral group is at least 6, they are all contained in the subgroup of rotations, and so the only way you can have a direct product is if that subgroup of rotations has order 4k+2, so that its Sylow 2-subgroup is centralized by a flip.