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

I have no idea at all how to attempt such question. I have read some solutions in which we find the cardinality of $kerm\phi$ for some $\phi mapping $ that is onto using the first theorem of homomorphism.

After that I don't understand anything

Is there a general approach to these questions ?

Arthur
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So Lo
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2 Answers2

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If there were a homomorphism, the kernel would be a subgroup of order $2$, since $Z_4 \oplus Z_4$ has order $16$, and the first isomorphism theorem implies that $Z_{16} \oplus Z_2/ker \phi \cong Z_4 \oplus Z_4$ There is only one subgroup of order $2$ in this group, and its quotient is not isomorphic to $Z_4 \oplus Z_4$.

Andres Mejia
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    can we say that $|ker \phi = 2|$ but clearly $\phi(0,0)= \phi(8,0) = \phi(4,0) = (0,0)$ which is the identity of $Z_4 \oplus Z_4$ Hence kernel size is greater than 2 and therefore $\phi$ can not be onto – So Lo Nov 24 '17 at 21:14
  • Sounds great${}{}$ – Andres Mejia Nov 24 '17 at 22:54
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Write $a$ and $b$ for the generators of the $Z_{16}$ and $Z_2$ on the left. If $\phi$ is a homomorphism from $Z_{16}\oplus Z_2$ to $Z_{4}\oplus Z_4$ then $\phi(4a)=0$. This means that the kernel of $\phi$ has size at least $4$. Can $\phi$ be onto with a kernel that big?

Angina Seng
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    can we say that $|ker \phi = 2|$ but clearly $\phi(0,0)= \phi(8,0) = \phi(4,0) = (0,0)$ which is the identity of $Z_4 \oplus Z_4$ Hence kernel size is greater than 2 and therefore $\phi$ can not be onto – So Lo Nov 24 '17 at 21:14
  • @SoLo Exactly!${}$ – Angina Seng Nov 24 '17 at 21:15