Find an isomorphism between U_7 and Z_7

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Homework Statement



There is an isomorphism of U_{7} with Z_{7} in which \zeta=e^{(i2\pi}/7\leftrightarrow4. Find the element in Z_{7} to which \zeta^{m} must correspond for m=0,2,3,4,5, and 6.

Homework Equations





The Attempt at a Solution


\zeta^{0}=0
\zeta^{2}=4+_{7}4=1
\zeta^{3}=\zeta^{2}\zeta^{1}=4+_{7}1=-2

I don't know if I'm doing this right?
 
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I would like to clarify a few things about your question. First, U_{7} and Z_{7} refer to the units and integers modulo 7, respectively. Isomorphism means that there exists a one-to-one mapping between the elements of these two groups that preserves their structure and operations. Therefore, the element \zeta in U_{7} corresponds to the integer 4 in Z_{7} in this specific isomorphism.

Now, for the element \zeta^{m}, we can use the property of isomorphism to find its corresponding element in Z_{7}. This property states that for any elements a and b in a group, their isomorphism will satisfy a^{m}\leftrightarrow b^{m}. So for m=0, we have \zeta^{0}=1 in U_{7}, which corresponds to 4^{0}=1 in Z_{7}. Similarly, for m=2, we have \zeta^{2}=1 in U_{7}, which corresponds to 4^{2}=2 in Z_{7}. For m=3, we have \zeta^{3}=\zeta^{2}\zeta^{1}=1\cdot\zeta=4 in U_{7}, which corresponds to 4^{3}=1 in Z_{7}. The same logic can be applied to find the corresponding elements for m=4,5, and 6.

I hope this clarifies your doubts and helps you solve the problem. Remember, isomorphism is a powerful tool that allows us to understand and compare different mathematical structures. Keep exploring and learning!
 
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