Find an isomorphism from Z_12 to Z_4 (direct product) Z_3. (Z_4 circle plus Z_3)

how many isomorphism are there in total.

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- Nov 8th 2008, 01:44 PMJuancd08Isomorphism
Find an isomorphism from Z_12 to Z_4 (direct product) Z_3. (Z_4 circle plus Z_3)

how many isomorphism are there in total. - Nov 8th 2008, 02:24 PMThePerfectHacker
If $\displaystyle \theta : \mathbb{Z}_{12} \to \mathbb{Z}_4 \times \mathbb{Z}_3$ is a homomorphism then $\displaystyle \theta (n) = \theta (1+...+1) = n\cdot \theta (1)$ (where $\displaystyle 0\leq n\leq 11$).

Now if $\displaystyle \theta$ is an isomorphism then $\displaystyle | 1| = |\theta(1)| = 12$. Thus, we require that order of $\displaystyle \theta (1)$ to be equal to $\displaystyle 12$. The elements that have order $\displaystyle 12$ in $\displaystyle \mathbb{Z}_4 \times \mathbb{Z}_3$ are $\displaystyle (1,1),(1,2),(3,1),(3,2)$. Now confirm that $\displaystyle \phi (1) = (1,1),(1,2),(1,3),(3,2)$ in each of four cases extends to an isomorphism. Thus, that means there are four isomorphisms.

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Here is another way to do this problem. Note that the number of isomorphisms between is the same as the number of automophisms of $\displaystyle \mathbb{Z}_n$. Now use the result that $\displaystyle |\text{Aut}(\mathbb{Z}_n)| = \phi (n)$. In this case $\displaystyle \phi(12) = \phi(4)\phi(3) = 4$. - Nov 8th 2008, 06:48 PMJuancd08order?
how do you get that those elements have order 12