Isomorphism

**Juancd08** · November 8th 2008, 01:44 PM

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.

**ThePerfectHacker** · November 8th 2008, 02:24 PM

Originally Posted by Juancd08

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

If $\theta : \mathbb{Z}_{12} \to \mathbb{Z}_4 \times \mathbb{Z}_3$ is a homomorphism then $\theta (n) = \theta (1+...+1) = n\cdot \theta (1)$ (where $0\leq n\leq 11$ ).

Now if $\theta$ is an isomorphism then $| 1| = |\theta(1)| = 12$ . Thus, we require that order of $\theta (1)$ to be equal to $12$ . The elements that have order $12$ in $\mathbb{Z}_4 \times \mathbb{Z}_3$ are $(1,1),(1,2),(3,1),(3,2)$ . Now confirm that $\phi (1) = (1,1),(1,2),(1,3),(3,2)$ 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 $\mathbb{Z}_n$ . Now use the result that $|\text{Aut}(\mathbb{Z}_n)| = \phi (n)$ . In this case $\phi(12) = \phi(4)\phi(3) = 4$ .

**Juancd08** · November 8th 2008, 06:48 PM

how do you get that those elements have order 12

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