Prove that there is no homomorphism from $Z_{16} \oplus Z_2$ onto $Z_4 \oplus Z_4$

Prove that there is no homomorphism from $Z_{16} \oplus Z_2$ onto $Z_4 \oplus Z_4$

Attempt: Suppose these exists a homomorphism $\Psi : (Z_{16} \oplus Z_2) \rightarrow (Z_4 \oplus Z_4)$

Then, since $\Psi$ is an onto homomorphism, this means :

$|Ker ~\Psi|= |Z_{16} \oplus Z_2|/|Z_4 \oplus Z_4| = 32/16 =2 ..........(1)$

(For any homomorphism $\Psi$ under additive binary composition, $\Psi$

But the following elements in the Set : $\{(0,0),(4,0),(8,0),(12,0)\}$ form a group $\subset Ker ~\Psi$ and order of this group is $4 \neq 2$ contrary to the conclusion in $(1)$

{This is because in case of groups under addition , if $\Psi$ be a homomorphism, then $\Psi(k) = k \Psi(1)$. Whenever $k$ is a multiple of $4$, in $Z_4 \oplus Z_4, \Psi[(k,0)]$ will reduce to $(0,0)$} Hence $\Psi$ cannot be a homomorphism.

Is this attempt correct?

• @MarcinŁoś and Michael : But, $\{(0,0),(4,0),(8,0),(12,0)\}$ forms a group and $\subset$ $Ker ~\Psi$ . But, this can't be possible because $|Ker ~\Psi|=2$ – MathMan May 26 '14 at 7:43
• Why is it contained in the kernel? – Michael Albanese May 26 '14 at 7:45
• Well, if you can show this, then that's it. But it seems you have just stated it. Perhaps it's obvious, but I cannot see it immediately. – Marcin Łoś May 26 '14 at 7:47
• That's because in case of groups under addition , if $\Psi$ be a homomorphism, then $\Psi(k) = k \Psi(1)$. Whenever $k$ is a multiple of $4$, in $Z_4 \oplus Z_4, \Psi[(k,0)]$ will reduce to $(0,0)$ – MathMan May 26 '14 at 7:50
• Yes. This is correct. Do make it clear (in the solution that you present to your teacher/students) that you are using the fact that if $f:G\to G'$ is a homomorphism of abelian groups such that $ny=0$ for all $y\in G'$ for some fixed integer $n$, then $nG=\{nx\mid x\in G\}$ is automatically a subgroup of the kernel. I see that you edited just that in while I was thinking/typing :-) – Jyrki Lahtonen May 26 '14 at 7:58