# 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?

• It is indeed correct. – Frieder Jäckel Oct 27 '17 at 16:21
• @FriederJäckel do you think there is any more to add or does this fully prove the two are isomorphic? – K Math Oct 27 '17 at 16:22
• There's nothing more to add because the image is all of $\mathbb{Z}_8.$ – Frieder Jäckel Oct 27 '17 at 16:24
• .. or show that the kernel of the canonical $\Bbb Z\to \Bbb Z_{24}\to \Bbb Z_{24}/\langle 8\rangle$ is precisely $8\Bbb Z$. – Hagen von Eitzen Oct 27 '17 at 16:30
• "Thus $\mathbb Z_{24}/\langle 8 \rangle$ is a homomorphism to $\mathbb Z_8$" is incorrect. $\mathbb Z_{24}/\langle 8 \rangle$ is a group, not a homomorphism. Also, you haven't shown that $f$ is a homomorphism, you merely claimed it. – Bungo Oct 27 '17 at 16:38

• 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.
• I know we typically prove a homomorphism with $f(a)f(b)=f(ab)$ but I don't know how to show this in this instance? – K Math Oct 30 '17 at 15:13
• It's easiest if you recall that the elements of $\mathbb{Z}_n$ are really equivalence classes modulo $n$. i.e. they are cosets of the form $a + n\mathbb{Z}$. Looking at it this way, the homomorphism you have defined is $\phi(a + 24\mathbb{Z}) = a + 8\mathbb{Z}$. So you have $\phi\left[ (a+b) + 24\mathbb{Z} \right] = (a+b) + 8\mathbb{Z} = \left(a + 8\mathbb{Z}\right) + \left(b + 8\mathbb{Z}\right)$. This last identity follows because of how addition of cosets works. – wgrenard Oct 30 '17 at 16:00