Direct product representation of $O_3$ contradicting the fact tht it is not abelian. I recently proved that
$$
O_3 \cong SO_3 \times C_2.
$$
Both elements in the direct product are abelian, and hence $O_3$ is abelian. This certainly can not be true. I don't understand where the error occurs. Surely the fact
"A group is abelian if and only if a direct product representation of it is abelian (given it exists) "
is right?
 A: The "direct product" formula $O_3 \cong SO_3 \times C_2$ is not correct. It is instead a semidirect product $O_3 \cong SO_3 \rtimes C_2$. One way to show this is to check directly that the map $(x,y,z) \mapsto (x,-y,z)$, a reflection across the $xz$-plane, which is an element of $O_3$ but not of $SO_3$, does not commute with the map $(x,y,z) \mapsto (-y,x,z)$, an element of $SO_3$ which rotates by $90^\circ$ around the $z$-axis.
Even one dimension down, where $SO_2$ is actually an abelian group, the group $O_2$ is not abelian, and there is no direct product formula $O_2 \cong SO_2 \times C_2$, instead there is a nontrivial semidirect product formula $O_2 \cong SO_2 \rtimes C_2$, and in particular the map $(x,y) \mapsto (x,-y)$, a reflection of the $xy$-plane across the $x$-axis, does not commute with the map $(x,y) \mapsto (-y,x)$, a $90^\circ$ rotation of the plane.
A: Something is not quite right. Indeed, the group $SO_3$ is not abelian, as rotations about different axes generally do not commute.
Also, you are correct: the direct product of abelian groups is again abelian.
