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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?

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  • $\begingroup$ Does the trivial homomorphism counts? I mean: Let $G$ = $\mathbb{Z}_{8} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ and $H$ = $\mathbb{Z}_{4} \times \mathbb{Z}_{4}$. Let $f: G \longrightarrow H$ be a function such that: $\forall g \in G, f(g) = e$, for $e$ the neutral element in $H$. Isn't it an homomorphism? $\endgroup$ – Diego Robayo Nov 28 '14 at 1:58
  • $\begingroup$ @D.A.Robayo The question asks for onto homomorphism. That's why $ker\phi\cong\Bbb Z_2$. $\endgroup$ – Quang Hoang Nov 28 '14 at 2:25
  • $\begingroup$ Misread, sorry. $\endgroup$ – Diego Robayo Nov 28 '14 at 3:56
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    $\begingroup$ the problem with your argument is that ker $\phi$ may sits inside $\mathbb{Z}_8.$ for example, ker $\phi$ may be equal to $\{(0, 0, 0), (4, 0, 0) \}.$ in this case the corresponding quotient doesn't have any element of order $8$. use the hint given by user lhf. that should be sufficient. $\endgroup$ – Krish Nov 28 '14 at 4:41
  • $\begingroup$ @Krish but wouldn't $(1,0,0) + \{(0,0,0),(4,0,0)\} = \{(1,0,0),(5,0,0)\}$, and both those elements have order 8? $\endgroup$ – user196009 Nov 28 '14 at 6:12
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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 $2$) $\Rightarrow (\mathbb{Z_8} \times \mathbb{Z_2} \times \mathbb{Z_2})/H \cong \mathbb{Z_4} \times \mathbb{Z_2} \times \mathbb{Z_2} \ncong \mathbb{Z_4} \times \mathbb{Z_4}.$

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  • $\begingroup$ I'm not sure why $G/H \cong \mathbb{Z}_4 \times \mathbb{Z}_2$. Shouldn't $G/H$ have order 16? $\endgroup$ – user196009 Nov 28 '14 at 7:46
  • $\begingroup$ I'm also confused on the isomorphisms. Why does $\mathbb{Z}_8 /<4> \cong \mathbb{Z}_4$ imply that $G/H \cong \mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$? $\endgroup$ – user196009 Nov 28 '14 at 8:05
  • $\begingroup$ @user196009: yeah, I wrote wrongly. I'll edit my answer. also see that for the case $H = \left<(4, 0, 0) \right>,$ $G/H \cong (\mathbb{Z}_8 /<4>) \times \mathbb{Z}_2 \times \mathbb{Z}_2.$ $\endgroup$ – Krish Nov 28 '14 at 8:24
  • $\begingroup$ @user196009: I gave a new argument. I hope it will be helpful. $\endgroup$ – Krish Nov 28 '14 at 8:50
  • $\begingroup$ In regards to the original argument, wouldn't $(1,0,0) + <(4,1,1)>$ have order 8, so the original idea still works? Wouldn't this be true for every subgroup except $<(4,0,0)>$? Also I'm not sure I quite understand, sorry. Why can we say that $G/H \cong (\mathbb{Z}_8/<4>) \times \mathbb{Z}_2 \times \mathbb{Z}_2$? Is it because for any $g = (x,y,z) \in G$ when we add it to $H$, only the first component will change? $\endgroup$ – user196009 Nov 28 '14 at 8:55
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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.

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