let $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\;J = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \; U = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \; V = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. these matrices together with their additive inverses give a faithful representation of the dihedral group $D_8$. $$ \\ $$ we may use the Kronecker product of pairs of these matrices, defined thus: $$ I \otimes U =\begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{pmatrix} \\ $$ whereas: $$ U \otimes I =\begin{pmatrix} U & 0 \\ 0 & U \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix} $$ denoting, as usual, the real quaternions by the symbol $\mathbb{H}$, the Kronecker products of the $D_8$ representation matrices re-appear when we describe the action of $\mathbb{H} \otimes \mathbb{H}^{\text{op}}$ on $\mathbb{H}$ defined by: $$ (a \otimes b)(h) = ahb $$ in terms of the usual basis $1,i,j,k$ for $\mathbb{H}$ this $\mathbb{R}$-module endomorphism of $\mathbb{H}$ corresponding to $(a \otimes b)$ can be expressed by a matrix $M_{a \otimes b} \in M_4(\mathbb{R})$ for which: $$ \forall h \in \mathbb{H}, M_{a \otimes b}h=ahb $$ to construct these matrices we extend by linearity from the correspondences in the following table (the 'additive inverses'are omitted for brevity). $$ \begin{matrix} \otimes & 1 & i & j & k \\ 1 & I \otimes I & V \otimes J & I \otimes V & V \otimes U \\ i & V \otimes I & -I \otimes J & V \otimes V & -I \otimes U \\ j & J \otimes V & -U \otimes U & -J \otimes I & U \otimes J \\ k & U \otimes V & J \otimes U & -U \otimes I & -J \otimes J \\ \end{matrix} $$ although the pattern is not simple to apprehend, test computations suggest it respects multiplication. for example, using componentwise quaternion multiplication - reversed for the second component $\in \mathbb{H}^{\text{op}}$: $$ (j \otimes i)(k \otimes j) = (jk \otimes ji) = -i \otimes k $$ using the table, this corresponds to: $$ (-U \otimes U)(-U \times I) = (-U)(-U) \otimes UI = I \otimes U $$
using the table 'in reverse', we see that $I \otimes U$ corresponds to $-i \otimes k$.
my question is twofold. firstly does the arithmetic check out? secondly, letting $\mathbb{Q_8}$ denote the quaternion group, does the table define an isomorphism between two groups of order 32, which i tentatively refer to as $\mathbb{Q_8} \otimes \mathbb{Q_8}^{\text{op}}$ and $D_8 \otimes D_8$?
will be grateful for any corrections, and/or contextualizations of the above observations, which i find puzzling.
Every group $(G,\circ)$ is isomorphic to its opposite group $(G,\bullet)$, defined by $a\bullet b:=b\circ a$, via the reciprocal function $x^{-1}$ (which is the same function on the set $G$ regardless of which of the two operations we're considering). This can be checked by hand. The key is that inversion is an antiautomorphism, i.e. a map $\phi$ satisfying $\phi(ab)=\phi(b)\phi(a)$. Indeed, any ring antiautomorphism also defines an isomorphismn between a ring $R$ and its opposite ring $R^{\mathrm{op}}$. (Note not all rings have antiautomorphisms, and in particular not all rings are isomorphic to their opposites.)
Inversion also allows us to convert between left and right group actions. For example, if $x\cdot g$ is a right group action, then $g\circ x:=x\cdot g^{-1}$ defines a left group action. The same is true for ring antiautomorphisms of $R$ and converting between left and right module structures over $R$. The quaternions $\mathbb{H}$ have an antiautomorphism: conjugation satisfies $\overline{ab}=\overline{b}\,\overline{a}$.
Note $\mathbb{H}$ acts on itself (module structure) from both the left and the right, and these actions commute (the associative property $a(xb)=(ax)b$, aka a bimodule). As a result, we can interpret underlying space of $\mathbb{H}$ as a module over $\mathbb{H}\otimes\mathbb{H}$ (omitting the $\mathrm{op}$ designation on the second factor) with multiplication defined by $(a\otimes b)x:=ax\overline{b}$. Since $Q_8$ is a subgroup of $\mathbb{H}$, this turns $\mathbb{H}$ into a linear representation of $Q_8\times Q_8$. However, one easily checks it has nontrivial kernel $\mathbb{Z}_2$ because the element $(-1)\otimes(-1)$ acts trivially, so the action factors through the quotient group.
This yields (an isomorphic copy of) the central product $Q_8\times_{\mathbb{Z}_2}Q_8=(Q_8\times Q_8)/\langle (-1,-1)\rangle$.
Similarly, the dihedral group $D_8$ of order $8$ has an irreducible real representation $\mathbb{R}^2$ containing the unit square, which means $D_8\times D_8$ has an irreducible representation $\mathbb{R}^2\otimes\mathbb{R}^2$, but again the element $(-1,-1)$ generates the nontrivial kernel, so we actually have a real four-dimensional representation of the central product $D_8\times_{\mathbb{Z}_2}D_8$.
You are right that these are isomorphic groups (and equivalent representations), although I did not fully check your multiplication table. It is possible to give an account of "where" this isomorphism comes from.
The $\mathbb{H}\otimes\mathbb{H}$-module structure on $\mathbb{H}$, taking the obvious ordered basis $\{1,i,j,k\}$ of $\mathbb{H}$, induces an isomorphism $\mathbb{H}\otimes\mathbb{H}\to\mathbb{R}(4)$. I will use the notation $\mathbb{K}(n)$ for the algebra of $n\times n$ matrices over a field (or skew field) $\mathbb{K}$. This is a common abbreviation in contexts related to Clifford algebras (of which $\mathbb{H}$ and $\mathbb{R}(4)$ are examples). We also have $\mathbb{R}$-algebra isomorphisms $\mathbb{C}\otimes\mathbb{C}\cong\mathbb{R}^2=\mathbb{R}\oplus\mathbb{R}$ (exercise) and $\mathbb{C}\otimes\mathbb{H}\cong\mathbb{C}(2)$ (view $\mathbb{H}$ as a 2D left $\mathbb{C}$-vector space and multiply by conjugates of quaternions on the right).
The quaternions $\mathbb{H}$ are generated as an $\mathbb{R}$-algebra by two anticommuting square roots of negative one, $i$ and $j$. The multiplicative subgroup they generate is $Q_8$. The split quaternions $\mathbb{H}_{\mathrm{sp}}$ are generated as an $\mathbb{R}$-algebra by an anticommuting square root of one and negative one, let's say $r^2=-1$ and $f^2=1$ satisfying $rf=-fr$. Since $-r=r^{-1}$ and $f=f^{-1}$, we can rewrite this last condition as $frf^{-1}=r^{-1}$, so this is exactly the presentation of the dihedral group. That is, the multiplicative subgroup of $\mathbb{H}_{\mathrm{sp}}$ generated by $r$ and $f$ is $D_8$.
It turns out, $\mathbb{H}_{\mathrm{sp}}\cong\mathbb{R}(2)$ - indeed, the elements $r$ and $f$ correspond to the rotation $[\begin{smallmatrix}0&-1\\1&\,0\end{smallmatrix}]$ and the reflection $[\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}]$ respectively. Another key idea is $\mathbb{R}(m)\otimes\mathbb{R}(n)\cong\mathbb{R}(mn)$. One explicit isomorphism is taking a tensor product to the corresponding Kronecker product of matrices.
Putting everything together, there is an $\mathbb{R}$-algebra isomorphism
$$ \mathbb{H}\otimes\mathbb{H}\to\mathbb{R}(4)\to\mathbb{R}(2)\otimes\mathbb{R}(2)\to\mathbb{H}_{\mathrm{sp}}\otimes\mathbb{H}_{\mathrm{sp}} $$
which one checks restricts to a group isomorphism
$$ Q_8\times_{\mathbb{Z}_2}Q_8\to D_8\times_{\mathbb{Z}_2}D_8. $$
For example, one can chase the following elements
$$ i\otimes 1\mapsto \left[\begin{smallmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{smallmatrix}\right]\mapsto[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}]\otimes[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}] \mapsto 1\otimes r $$
$$ 1\otimes j\mapsto \left[\begin{smallmatrix}0&0&-1&0\\0&0&0&-1\\1&0&0&0\\0&1&0&0\end{smallmatrix}\right]\mapsto[\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}]\otimes[\begin{smallmatrix}1&0\\0&1\end{smallmatrix}]\mapsto r\otimes1 $$