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Show that the multiplicative group $\mathbb{Z}_{10}^{\times}$ is isomorphic to the additive group $\mathbb{Z}_4$.

I'm completely lost with this one.

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First off, we have $\mathbb{Z}_4^+=\{0,1,2,3\}$, and $\mathbb{Z}_{10}^\times = \{1,3,7,9\}$.

Now, I'm going to tell you that $f:\mathbb{Z}_4^+ \to \mathbb{Z}_{10}^\times$ is a homomorphism for which $f(1) = 3$. What you have to figure out is: what are the values of $f(2),f(3),$ and $f(4)$, and why is this an isomorphism?


At this point, you should have the following: $$ \begin{align} f(1) &= 3\\ f(2) &= f(1+1) = f(1)f(1) = f(1)^2 = 9\\ f(3) &= \cdots = f(1)^3=7\\ f(4) &= \cdots = f(1)^4 = 1 \end{align} $$ In other words, we've defined $f$ by $f(n) = 3^n \pmod{10}$ for $n = 0,1,2,3$. Now, to prove $f$ is a homomorphism, it suffices to state that $$ f(n+m) = 3^{n+m}=3^n\cdot3^m \pmod{10} = f(n)f(m) $$

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  • $\begingroup$ So $f(2)=7$, $f(3)=9$, & $f(4)=1$. Why can you say $f$ is a homomorphism? $\endgroup$ – Desperate Fluffy Nov 27 '13 at 7:28
  • $\begingroup$ Not quite. $$f(2) = f(1+1) = f(1)f(1) = 3\times3 = 9$$ Can you take it from there? $\endgroup$ – Omnomnomnom Nov 27 '13 at 7:30
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    $\begingroup$ I think so. Thank you! ^_^ $\endgroup$ – Desperate Fluffy Nov 27 '13 at 7:30
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write down all the elements of $\mathbb{Z}^{\times}_{10}$ explicitly. any find a generator by its Cayley Table or by explicit calculation. Now map this generator to $1$ in $\mathbb{Z}_4$.

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  • $\begingroup$ Its Cayley Table? $\endgroup$ – Desperate Fluffy Nov 27 '13 at 7:08
  • $\begingroup$ ya. I mean you can use it to find the power an element. $\endgroup$ – GA316 Nov 27 '13 at 8:16

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