Prove $\text{Inn}(\mathbb{Q_8}) \cong \mathbb{Q_8}/ Z(\mathbb{Q_8})$

Question

I found on wikipedia somewhere that $\text{Inn}(\mathbb{Q_8}) \cong \mathbb{Q_8}/ Z(\mathbb{Q_8})$ but I don't know how to prove that their orders are equal.

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Also, note: $Z(\mathbb{Q_8}) = \{-1,1\}$ so $| \mathbb{Q_8}/ Z(\mathbb{Q_8}) |= \frac{| \mathbb{Q_8}|}{2} = 8/2=4$

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How do I prove $|\text{Inn}(\mathbb{Q_8}) | = 4$?

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Note: $\mathbb{Q_8}$ is the quaternion group = $\{-1, 1, -i, i, -j, j, -k, k\}$

Answer 1

We have, in general, that the canonical homomorphism $G\to \operatorname{Inn}(G)$

1. is surjective
2. has $Z(G)$ as kernel

Both of these are nice exercises to work through. And I don't think there is an easier way to show that the orders of $\operatorname{Inn}(G)$ and $G/Z(G)$ are equal than to use these two facts to show that they are, in fact, isomorphic.