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Prove that there is no surjective homomorphism from $\Bbb Z_{16} \oplus \Bbb Z_{2}$ onto $\Bbb Z_4 \oplus \Bbb Z_4$.

The book I'm reading starts off the proof with:

Suppose $\phi$ is such a homomorphism between the two groups, then $f: \Bbb Z_{16} \oplus \Bbb Z_{2} / \ker \phi \rightarrow \phi(\Bbb Z_{16} \oplus \Bbb Z_{2})$ is an isomorphism. Therefore $\ker \phi = \langle (8,1) \rangle, \langle (0,1 \rangle),$ or $\langle (8,0)\rangle$.

How does the isomorphism imply that this is what the kernel equals?

Oliver G
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3 Answers3

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The order of the domain is twice that of the range, so if the homomorphism is surjective, then the kernel has order $2$. That last sentence just listed all of the subgroups of the domain of order $2$.

florence
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Note that two isomorphic groups have the same number of elements. So, $|\Bbb Z_{16} \oplus \Bbb Z_2/\ker \phi| = 16$. So, $|\Bbb Z_{16}\oplus \Bbb Z_2|/|\ker \phi| = 16$.

So, $|\ker \phi| = 2$.

Ben Grossmann
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  • How did you get $|\Bbb Z_{16} \oplus \Bbb Z_2/\ker \phi| = 8$? – Oliver G Aug 24 '16 at 20:28
  • Excuse me, that should have been $16$. Fixed it. – Ben Grossmann Aug 24 '16 at 20:31
  • I guess I mean how did you find the quotient for the order of the factor group? Wouldn't it depend on what $\ker \phi$ is? – Oliver G Aug 24 '16 at 20:33
  • I found $|\Bbb Z_4 \oplus \Bbb Z_4| = 16$, hence the first equation. After I split up the quotient using Lagrange's theorem, I used the fact that $|\Bbb Z_{16} \oplus \Bbb Z_2| = 32$. – Ben Grossmann Aug 24 '16 at 20:36
  • So for $H$ a subgroup of $G$, $|G/H| = |G:H| = \frac{|G|}{|H|}$? Meaning the index of a subgroup equals the order of the factor group of the subgroup? – Oliver G Aug 24 '16 at 20:38
  • Yes, exactly! Remember that the elements of a factor group are cosets, which are precisely what the index counts. – Ben Grossmann Aug 24 '16 at 20:41
  • That makes sense, but why does the fact that the kernel only has two elements imply that these are the only possibilities? I can see that using other subgroups increase the size of the kernel, but does this property imply because it's of order 2 or is this because of the nature of the particular sets used? – Oliver G Aug 24 '16 at 20:50
  • The groups listed are the only groups of order $2$. – Ben Grossmann Aug 24 '16 at 21:07
  • I meant the only subgroups of order $2$, excuse me. – Ben Grossmann Aug 24 '16 at 22:06
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The cardinality of $Z_{16}\times Z_2$ is 32 and the cardinality of $Z_4\times Z_4$ is 16. Thus the cardinality of the kernel of $\phi$ is 2 since it is surjective. The only subgroups of order 2 of $Z_{16}\times Z_2$ are generated by $(8,1), (0,1), (8,0)$.

Tsemo Aristide
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