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A group homomorphism sends identity to identity. For $\theta (g),\theta (h)$ in its image,which is still in the image; hence the image of a group homomorphism is a subgroup of $H$. Let $N=\ker \theta$. It is a subgroup by the same identity, and for $k\in N$ and $g\in G$,Thus $gN{g}^{-1}\subseteq N$, and replacing $g$ by $g^{-1}$ gives equality, proving that $N$ is a normal subgroup.

Define a map from the quotient group byIf $gN=hN$, then $h^{-1}g\in N$ and $\theta (g)=\theta (h)$, so the map is well defined. Conversely equality of the images implies $h^{-1}g\in N$, giving injectivity. It is onto by the definition of image, and $\Phi ((gN)(hN))=\theta (gh)=\theta (g)\theta (h)$. This proves the first isomorphism theorem for groups in this case:An orthogonal matrix $Q$ has $Q^{\mathsf{T}}Q=I$, so $(\det Q{)}^2=1$. The determinant is therefore a group homomorphism $O(3)\to \{1,-1\}$, with image both signs because $\mathrm{diag}(-1,1,1)$ has determinant $-1$. Its kernel is the special orthogonal group $SO(3)$. Hence $SO(3)$ is normal in $O(3)$, andFor the last request, the determinant-twist projection in odd dimension gives an explicit example:Indeed $\det \pi (Q)=(\det Q{)}^4=1$, and scalar factors commute, so $\pi (QR)=\pi (Q)\pi (R)$. If $\pi (Q)=I$, then $Q=(\det Q)I$, making $Q=I$ or $Q=-I$. Both lie in the kernel, which is therefore exactly $\{I,-I\}$, of order two. The map also fixes every member of $SO(3)$, so it is onto.