can you reduce the order of an equality expression with infinite groups?

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I have $G \cong R^3$ and $Z(G) \cong \mathbb{R}$ (the whole process up until here is omitted for shortness)

It can be easily shown that $G \cong \mathbb{R}^3$ thus $|G| = |\mathbb{R}^3|$

We know that $|G| = |G/Z(G)||Z(G)|$ and we showed that $|Z(G)| = |\mathbb{R}|$

Thus $|\mathbb{R}^3|=|G/Z(G)||\mathbb{R}|$ so it must be that $|G/Z(G)| = |\mathbb{R}^2|$

I am not sure if the above is correct, and if it is I would like to know at leas the intuition of how to prove it (the result is correct, the method may be wrong)