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There are nontrivial proper normal subgroups. The determinant defines a surjective group homomorphism $G\to \{1,-1\}$. Its kernel is the rotational symmetry group of a cube, a normal subgroup of order twenty-four. Also $\{I,-I\}$ is a central subgroup of order two: central inversion preserves the cube and commutes with every matrix. Both provide affirmative examples.

For the body diagonal through $(1,1,1)$, write a general cube symmetry as $D_{\epsilon }P$, with $P$ a permutation matrix and $D_{\epsilon }=\mathrm{diag}({\epsilon }_1,{\epsilon }_2,{\epsilon }_3)$, each ${\epsilon }_j=\pm 1$. Since $P(1,1,1{)}^T=(1,1,1{)}^T$, preserving the unoriented diagonal requires $({\epsilon }_1,{\epsilon }_2,{\epsilon }_3)$ to be either $(1,1,1)$ or $(-1,-1,-1)$. ThusThe map $(\pi ,\epsilon )\mapsto \epsilon P_{\pi }$ from $S_3\times C_2$ to $H$ is a group homomorphism, because central inversion commutes with every coordinate permutation. It is surjective by the displayed description and injective because a permutation matrix cannot equal the negative of another permutation matrix. Consequently the body-diagonal stabilizer in the cube symmetry group isThis also agrees with the orbit-stabilizer theorem for the transitive action on the four body diagonals. The line is preserved as a set, so exchanging its opposite endpoints is allowed.