Is proving $O(3)\cong \mathbb{Z}_2 \otimes SO(3)$ equivalent to proving $O(3)/SO(3) \cong \mathbb{Z}_2$ I think proving the later is easier, see my other question, $O(3)/SO(3) \cong \mathbb{Z}_2$.
If given the statement, "Show the isomorphism $O(3)\cong \mathbb{Z}_2 \otimes SO(3)$" on an exam, is it legitamte to prove $O(3)/SO(3) \cong \mathbb{Z}_2$ instead? And why? Why is dividing/multiplying on each side of the equation like these are just numbers valid?
 A: As commenters point out, the notation $\otimes$ is not used for groups, only for vectorspaces. What you are thinking about is the direct product $\times$.
For groups $G$ and $H$ we have that the group $G \times H$ consists of all pairs $(g, h)$ with $g \in G$ and $h \in H$ where the group operation is just 'pointwise': for multiplicative groups we have $(g_1, h_1)(g_2, h_2) = (g_1g_2, h_1h_2)$ and for additive groups we have $(g_1, h_1) + (g_2, h_2) = (g_1 + g_2, h_1 + h_2)$. So everything happens inside the separate copies, there is no interaction between the groups so to speak.
There are some subtleties. We have that $\mathbb{Z}_2 \times \mathbb{Z}_3 \cong \mathbb{Z}_6$ but we do not have that $\mathbb{Z}_2 \times \mathbb{Z}_2 \cong \mathbb{Z}_4$.
This latter example is an answer (of the 'no: see this counter-example' type) to your question because we do have that $\mathbb{Z}_4/\mathbb{Z}_2 \cong \mathbb{Z}_2$.
I recommend you write out these really small examples explicitly to understand what is going on here.
The 'four maps' I was talking about in my answer to you other question are the following.
There are two 'outgoing' homomorphisms $G \times H \to G$ and $G \times H \to H$ given by $(g, h) \mapsto g$ and $(g, h) \mapsto h$. Also there are two 'incoming' maps: one map $G \to G \times H$ given by $g \mapsto (g, e)$ where $e$ is the neutral element of $H$ and one $H \to G \times H$ by $h \mapsto (e', h)$ where $e'$ is the neutral element of $G$.
Writing $B$ (for big) for the big group $G \times H$ we have a copy of $G$ sitting inside $B$ as $(G, e)$ (i.e. the image of the incoming map from $G$). This copy of $G$ is a normal subgroup so there exist a quotient $B/G$. It is not hard to see that this quotient must be isomorphic to $H$, and in a sloppy sense the outgoing map to $H$ gives you the isomorphism.
However: in the general situation that you have a normal subgroup isomorphic to $G$ in a big group $B$ such that $B/G \cong H$ we do not have that $B = G \times H$. For that to be true we need that moreover we have a normal subgroup isomorphic to $H$ sitting inside $B$ such that the quotient $B/H$ (that then apparently also exists) is isomorphic to $G$. In other words: you need to do twice as many work to show $B = G \times H$ than to show that $B/G = H$.
I hope this helps.
