Can $ \frac{\mathbb{Z_{4} \times \mathbb{Z_{6}}}}{\left<(0,2)\right>} $ be isomorphic to $ \mathbb{Z_{8}} $?

Question

This is the example 15.10 of Fraleigh's textbook:

Compute the factor group $$ \frac{\mathbb{Z_{4} \times \mathbb{Z_{6}}}}{\left<(0,2)\right>}.$$

$ \mathbb{Z_{4} \times \mathbb{Z_{6}}} $ has order $4\cdot 6 = 24$.
$(0,2)$ generates a subgroup of order 3, hence the order would be 8.
The book says it is isomorphic to $ \mathbb{Z_{4} \times \mathbb{Z_{2}}} $. I got this far and just curious if I can think the answer also can be $ \mathbb{Z_{8}} $.

Answer 1

Your answer is based on the structure of abelian groups with 8 elements, but this is not necessary in order to solve the problem. In fact, $\left<(0,2)\right>=\{0\}\times 2\mathbb Z_6 $, and then $$\frac{\mathbb Z_4 \times \mathbb Z_6}{\left<(0,2)\right>}=\frac{\mathbb Z_4 \times \mathbb Z_6}{\{0\}\times 2\mathbb Z_6}\simeq\frac{\mathbb Z_4}{\{0\}}\times\frac{\mathbb Z_6}{2\mathbb Z_6}\simeq\mathbb Z_4\times\mathbb Z_2.$$

Answer 2

The potential isomorphism of $\frac{\mathbb Z_4\times\mathbb Z_6}{\langle(0,2)\rangle}$ to $\mathbb Z_8$ will give rise to an "onto" homomorphism of $\mathbb Z_4\times\mathbb Z_6$ to $\mathbb Z_8$. In that homomorphism, the element of $\mathbb Z_4\times\mathbb Z_6$ that maps to a generator of $\mathbb Z_8$ would have an order divisible by $8$. However, orders of any elements in $\mathbb Z_4\times\mathbb Z_6$ are divisors of $\gcd(4,6)=12$, and no divisors of $12$ are divisible by $8$.