They are isomorphic

Prove that:

$$\mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_8 \bigg/\langle [1]_4, [2]_4, [4]_8 \rangle \cong \mathbb{Z}_4 \times \mathbb{Z}_8$$

Solution

We are ignoring $\mathbb{Z}_2$ and we are construncting the following trivial homomorphism:

$$ \varphi\left ( \left [ \kappa \right ]_4, \; \left [ \lambda \right ]_4 , \; \left [ \mu \right ]_8 \right ) = \left ( \left [ \lambda \right ]_4, \; \left [ \mu \right ]_8 \right )$$

The kernel of this $\varphi$ is exactly $[1]_4, [2]_4, [4]_8$, because $\kappa$ does not have any restriction , so $[1]_4$ is convinient and:

$$\varphi\left ( \left [ \kappa \right ]_4, \; \left [ \lambda \right ]_4 , \; \left [ \mu \right ]_8 \right ) =0 \Leftrightarrow \lambda=2=\mu$$

since they both get zero at the same time. It is rather easy to check that this is an $1-1$ depection as well as onto. Thus these two sets are isomorphic.