Prove that $\mathbb Z_8$ and $\mathbb Z_{24}/\langle 8\rangle$ are isomorphic. 
Use the Fundamental Homomorphism Theorem to prove that $\mathbb Z_8$ and $\mathbb Z_{24}/\langle 8\rangle$ are isomorphic.

The following function is a homomorphism from $\mathbb Z_{24}$ to $\mathbb Z_8$:
$\bigl(\begin{smallmatrix}
  0& 1 & 2 & 3 & 4 &5&6&7&8&9&10&11&12&13&14&1&16&17&18&19&20&21&22&23 \\
  0 & 1 & 2 &3 & 4&5&6&7&0&1&2&3&4&5&6&7&0&1&2&3&4&5&6&7
\end{smallmatrix}\bigr)$.
Thus we can say $\operatorname{ker}(f)=\{0,8,16\}$.
Thus $\mathbb Z_{24}/\langle 8\rangle$ is a homomorphism to $\mathbb Z_8$.
By the FHT, $\mathbb Z_8 \cong \mathbb Z_{24}/\langle 8\rangle$.
Is this the correct use of FHT? I thought I needed to show that group/ker $\cong$ im?

The general idea is correct. To use the first isomorphism theorem here you want to:


*

*Find a homomorphism $\phi:\mathbb{Z}_{24} \to \mathbb{Z}_8$

*Show that $\phi$ is surjective (i.e. $\text{Im}(\phi) = \mathbb{Z}_8$)

*Show that $\ker(\phi) = \langle 8 \rangle$


Then you may conclude that $\mathbb{Z}_{24}/\langle 8 \rangle \cong \mathbb{Z}_8$. In your proof, you claimed to have found the required homomorphism, but you still need to prove that it is indeed a homomorphism. This is the biggest missing link in your proof. Also, though it is obvious from the way you have written the map, it doesn't hurt to explicitly mention in your proof that $\phi$ is surjective, because this is indeed important.
Added: As for your last question. You don't need to show that $\mathbb{Z}_{24}/\ker(\phi) \cong \text{Im}(\phi)$. This is precisely the statement of the first isomorphism theorem. You are simply using this fact in your proof.
