There is no homomorphism from $\mathbb{Z_8} \times \mathbb{Z_2} \times \mathbb{Z_2}$ onto $\mathbb{Z_4} \times \mathbb{Z_4}$ If such a homomorphism $\phi$ existed, then the first isomorphism theorem says that $|\ker \phi| = 2$. Since $\mathbb{Z_8} \times \mathbb{Z_2} \times \mathbb{Z_2}$ is abelian, then every subgroup is normal, so $\ker \phi$ must be one of the subgroups of $\mathbb{Z_8} \times \mathbb{Z_2} \times \mathbb{Z_2}$ of order 2. Denote these subgroups by $H_1, ... , H_k$. Now $(1,0,0) + H_i$ for any $1 \leq i \leq k$ has an element of order $8$, but 4 is the largest order for any element in $\mathbb{Z_4} \times \mathbb{Z_4}$. Thus no homomorphism exists.
Is this proof OK?
 A: Your argument is not entirely correct (as it doesn't cover all the cases), but it can be changed slightly to get the exact contradiction. Let $G := \mathbb{Z_8} \times \mathbb{Z_2} \times \mathbb{Z_2}, H :=$ ker$\phi$. As you have noted $| H | = 2.$ So it's a cyclic group generated by an element of order $2$. Now what are all the order $2$ elements of $G$? You can check that in each of the cases, you will get some contradiction. For example, if $H = \left <(4, 0,0) \right>,$ then $G/H \cong \mathbb{Z_4} \times \mathbb{Z_2} \times \mathbb{Z_2}$ (this was discussed in the answer of johng). If $H = \left <(0, 1,0) \right>,$ then $G/H \cong \mathbb{Z_8} \times \mathbb{Z}_2,$ and in this case we get a contradiction by the argument you gave. But you need to be careful for the cases like $H = \left< (4, 1, 1)\right>,$ or $H = \left<(4, 1, 0)\right>$ etc. to find the isomorphic classes of $G/H.$
Since I couldn't come up with a satisfactory explanation of the above arguments,
 I am giving a new proof (see EDIT). But I'm keeping the old one also so that someone else might give a better explanation.
${\bf EDIT:}$ Suppose $\phi : \mathbb{Z_8} \times \mathbb{Z_2} \times \mathbb{Z_2} \rightarrow \mathbb{Z_4} \times \mathbb{Z_4}$ be a surjective homomorphism and $H :=$ ker$\phi.$ Then $|H| = 2.$ Since there is no element of order $8$ in $\mathbb{Z_4} \times \mathbb{Z_4},$ every order $8$ element of $\mathbb{Z_8} \times \mathbb{Z_2} \times \mathbb{Z_2}$ must map to elements of order $1, 2$ or $4$. Let $a = (1, 0, 0) \in \mathbb{Z_8} \times \mathbb{Z_2} \times \mathbb{Z_2}.$ Then $|\phi (a)| = 1, 2, 4.$ If $|\phi (a)| = 1,$ then $a \in H \Rightarrow |H| \geq 8.$ If $|\phi (a)| = 2,$  then $\phi(2a) = 2\phi(a) = 0 \Rightarrow 2a \in H \Rightarrow \left<2a\right> \in H \Rightarrow|H| \geq 4.$ If $|\phi (a)| = 4,$ then $\phi (4a) = 4\phi(a) = 0 \Rightarrow$
 $(4, 0, 0) \in H \Rightarrow H = \left<(4, 0, 0)\right> $ (since $H$ is generated by an element of order $2$) $\Rightarrow (\mathbb{Z_8} \times \mathbb{Z_2} \times \mathbb{Z_2})/H \cong \mathbb{Z_4} \times \mathbb{Z_2} \times \mathbb{Z_2} \ncong \mathbb{Z_4} \times \mathbb{Z_4}.$
A: You're on the right track.  If there is such a homomorphism, the kernel H has order 2.  Now (1,0,0)+H has order at most 4. (Max order of elements in the image).  So (4,0,0) is in H.  But (4,0,0) has order 2, so H=<(4,0,0)>. Thus for $G=Z_8\times Z_2\times Z_2$, G/H is isomorphic to $Z_4\times Z_2\times Z_2$ which is not isomorphic to the image.
