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For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours.

Prove that no homomorphism exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$.

Assume a homomorphism $\phi$ exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$. Then $\phi$ is an isomorphism.

Note that the $\left | \ker \phi \right |=2$ The $\ker \phi$ is also a normal subgroup for $Z_{16}\oplus Z_{2}$. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups.

By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements.

Hence, $\ker\phi$ = $\left \{ (0,0),(8,0) \right \}$ ,$\left \{ (0,0),(0,8) \right \}$,$\left \{ (0,0),(8,1) \right \}$, possibly.

By the First isomorphism theorem:

$\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2$

$\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2$

$g\ker \phi \mapsto \phi(g)=\Psi(g\ker \phi)$

Any help is appreciated.

Edit: I know that $(2,0)$ has order $8$ in $Z_{16}\oplus Z_{2}$ but any elements in $Z_{4}\oplus Z_{4}$ does not have order $8$ which would have solved the question at the outset. But I would like a different route.

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    $\begingroup$ As abelian group, why not check $Hom(Z_{16},Z_4),Hom(Z_2,Z_4)$ first? $Hom(Z_{16}\oplus Z_2,Z_4\oplus Z_4)\cong Hom(Z_{16},Z_4)^2\oplus Hom(Z_2,Z_4)^2$ $\endgroup$ – user45765 Jul 11 '17 at 0:46
  • $\begingroup$ Btw, you can realize them as $Z-$module which is even better. $\endgroup$ – user45765 Jul 11 '17 at 0:50
  • $\begingroup$ @RobertIsrael Because the cardinality of the groups are not preserved under the map and so not bijective and hence not surjective. That's fine. But what about using the first isomorphism theorem to go about this one? $\endgroup$ – Mathematicing Jul 11 '17 at 0:51
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    $\begingroup$ Why do you say that if a surjective map exists it is an isomorphism? The two groups have different orders... $\endgroup$ – Mariano Suárez-Álvarez Jul 11 '17 at 3:56
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    $\begingroup$ Also: since the groups are abelian,all subgroups are normal, be they of order $2$ or not. $\endgroup$ – Mariano Suárez-Álvarez Jul 11 '17 at 3:57
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Let $f:Z_{16}\times Z_2\to Z_4\oplus Z_4$ be a map. The subgroup $f(Z_{16})$ is cyclic, so it has at most $4$ elements (as $4$ is the maximum order of an element in the codomain of $f$) Now the image of $f$ is equal to the subgroup $f(Z_{16})+f(Z_2)$, which has at most $8$ elements. The map $f$ is therefore not surjective.

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At first see below to observe that, isomorphism $\phi$ preserves the order of an element:

Isomorphism $f$ preserves the order of an element?



Suppose on contrary that such a homomorphism exists, and consider the induced isomorphism $\Psi$, as you defined.

The group $\mathbb{Z}_4 \oplus \mathbb{Z}_4$ has exactly $12=16-4$ elements of order $4$; i.e. all elements which does not lie in the sub-group $\{0, 2\} \oplus \{0, 2\}$.

On the other hand $\mathbb{Z}_{16} \oplus \mathbb{Z}_2$ has only $6=8-2$ elements of order $4$; i.e. all elements of the sub-group $\{ 0, 4, 8, 12 \} \oplus \{ 0, 2 \}$ which does not lie in the sub-group $\{0\} \oplus \{0, 2\}$.



So the quoteint group appeared in the induced isomorphism $\Psi$ will only have at most $6$ elements of order $4$, which is a contradiction.

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