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Question about methods of finding homomorphisms from $\mathbb Z_4$ to $\mathbb Z_2 \times \mathbb Z_2$

I've seen both methods, but one seems to fail here. What went wrong?

  1. $\mathbb Z_4=\langle 1\rangle$, so homomorphism is defined by the value we give to $f(1)$. So we can give $4$ values, hence there are $4$ homomorphisms.

  2. We know $\ker f\trianglelefteq \mathbb Z_4$, so $|\ker f|$ is either $1,2,4$.

With this method, I can only find $2$ homomorphisms, trivial and another in which $|\ker f|=2$. What went wrong?

Thanks, Sammy

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There are two types of homomorphism - the trivial homomorphism, and one in which the element of order $4$ gets mapped to an element of order $2$.

This second type can be realised in three distinct ways, because there are three different elements of order $2$ which you can pick - identifying the kernel gives the isomorphism type of the image, but it doesn't tell you anything about the actual images of the elements. All three of these homomorphisms have the same kernel.

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  • $\begingroup$ So the second method can cause me "forgetting" some homomorphisms? $\endgroup$ – SammyBinZayyin May 6 '14 at 15:44
  • $\begingroup$ @SammyBinZayyin - it tells you that the images with that kernel are groups of order $2$ - then you have to go and see which groups of order $2$ are available. If you were mapping to a group of order $3$ or $5$ you'd have no homomorphisms of this type. As it is, since there are no elements of order $4$ in the image you've already excluded that possibility. Identifying the kernel and the isomorphism class of the potential image is a potential step on the route to a solution, in that it can help you to know where to look, but it doesn't solve the problem of identifying the concrete homomorphisms. $\endgroup$ – Mark Bennet May 6 '14 at 15:48

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