Symmetry subgroup of a cube

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Let $\epsilon=(\epsilon_1,\epsilon_2,\epsilon_3)$ with $\epsilon_i\in\{\pm 1\}$ and consider for each $\epsilon$:

$$ D_\epsilon=\begin{pmatrix} \epsilon_1 & 0 & 0\\ 0 & \epsilon_2 & 0\\ 0 & 0 & \epsilon_3 \end{pmatrix} $$

Let $O_h\subseteq GL_3(\mathbb{R})$ be

$$ O_h =\{P_\sigma D_\epsilon\in\mathbb{R}^{(3,3)}\mid \sigma\in S_3, \epsilon\text{ as above}\} $$

where $P_\sigma$ is the associated permutation matrix with $(P_\sigma)_{ij}=\delta_{i\sigma(j)}$. I want to prove that $O_h$ forms a subgroup of $GL_3(\mathbb{R})$. I've already had some thoughts about it, the cases for associativity is clear as it would be inherited from $GL_3(\mathbb{R})$, same goes for the identity and its properties. My problem actually lies with proving closure of the set. My guess is that for $A=P_\sigma D_\epsilon,B=P_\tau D_{\epsilon'}$, we have

$$AB=P_\sigma D_\epsilon P_\tau D_{\epsilon'}= P_{\sigma\circ\tau}D_{\epsilon''}$$

where $\epsilon''_i=\epsilon_{\sigma(i)}\epsilon'_i$.