Two centerless subgroups $H, K$ of $S_3\boxtimes C_2$, each of order six, such that $HK$ is not centerless. This is Exercise 3.2 of Roman's "Fundamentals of Group Theory: An Advanced Approach."
The Details:
Here $G_1\boxtimes G_2$ is the external direct product of $G_1$ and $G_2$, given by the Cartesian product of $G_1$ with $G_2$ under component-wise product.
A group $G$ is centerless if $Z(G)$ is trivial.
The set product of $H, K\le G$ is defined as $$HK=\{ hk\mid h\in H,k\in K\}.$$
The Question:

Let $G=S_3\boxtimes C_2$. Show that $G$ has two subgroups $H$ and $K$ that are centerless but that $G=HK$ is not centerless.

Thoughts:
I'm aware that $G$ is isomorphic to $D_6$, the dihedral group of order $12$, so it has the presentation
$$G\cong \langle a,b\mid a^6, b^2, bab=a^{-1}\rangle.$$
I'm not sure how that helps.
According to group names, there are three subgroups of $G$ of order six, one isomorphic to $C_6$, which is not centerless (since it is abelian), and two isomorphic to $S_3\cong D_3$.
One of the subgroups isomorphic to $S_3$ is given by
$$\{(r,[0]_2)\mid r\in S_3\} ;$$
however, I have a mental block as to what the other copy of $S_3$ is; is it
$$\{(s,[1]_2)\mid s\in S_3\} ?$$
The identity of $G$ is obtained by $(t, [1]_2)^2$ for $t$ of order two in $S_3$.
This is a question I ought to be able to solve by myself. Moreover, how might one approach this question without the help of group names or the knowledge that $G\cong D_6$? Presentations aren't covered in the book so far, too, so my initial observation is not in the spirit of the question.
I can't see the wood for the trees.
Please help :)
 A: You don't need to use $D_{6}$ (or $D_{12}$, depending on your naming scheme).
Note that $G$ is not centerless ($\{e\}\boxtimes C_2$ is the center), so if $HK=G$, you will get that it is not centerless.
So, your subgroups $H$ and $K$ should avoid the center. Also, they better be of order $6$, since any smaller order will be abelian. So you want two subgroups of order $6$, each isomorphic to $S_3$; their product should have order $12$, so you are going to want them to intersect in a subgroup of order $3$. That subgroup must project trivially onto the $C_2$ component, just by order considerations. And neither subgroup should contain an element of the form $(\sigma,e)$ and one of the form $(\sigma^{-1},x)$ (where $x$ is the nontrivial element of $C_2$, written multiplicatively).
So one subgroup can probably be $S_3\boxtimes \{e\}$. The other should intersect it in the $3$-subgroup of this subgroup. Then you want an element of order $2$ to finish generating $K$, but it shouldn't be in $H$...

Another way to think about it: imagine you are trying to teach someone about the sign of a permutation. To make sure they get it, you could "flag" every permutation with its sign: $-1$ if it is odd, $1$ if it is even. So the elements of $S_3$ would be $\mathrm{id}_1$, $(123)_1$, $(132)_1$, $(12)_{-1}$, $(13)_{-1}$, and $(23)_{-1}$. But, of course, once you know how to tell if a permutation is even or odd, you can drop the flag/subscript and get the exact same group...
A: How about two copies of $S_3$ that are subgroups?  $S_3$ is centerless.  Meanwhile $D_6$ is not centerless.
That  is, you stated, essentially, that there are two different subgroups isomorphic to $S_3$.  So you are essentially done, since, if we call them $H$ and $K$, we have $HK=G$ just by counting.
