It can be shown that the multiplicative group $(\mathbb{Z}_{10})^\times$ is cyclic. To which group is it isomorphic? It can be show that the multiplicative group $(\mathbb{Z}_{10})^\times$ is cyclic. To which group is it isomorphic?
What I have tried:
I have calculated the order of $(\mathbb{Z}_{10})^\times$ to be 4. Hence I think it is isomorphic to $\mathbb{Z}_{4}$.
Would someone be able to verify if my answer is correct?
Thank you for taking the time to respond to my question.
 A: Note that $(\mathbb{Z}/10\mathbb{Z}), \times)$ has 4 elements in it. So it is isomorphic to $(\mathbb{Z}/4\mathbb{Z}, +)$, indeed any cyclic group with 4 elements. Note that for each $m$ there is exactly one cyclic group of order $m$, up to group isomorphism.
Meanwhile to show that $(\mathbb{Z}/10\mathbb{Z}), \times)$ is indeed cyclic, note that 3 generates every element in this group.
A: You are correct that the order of $(\mathbb Z_{10})^\times$ is $4$,
and there is only one cyclic group of order $4$ up to isomorphism.
The set of elements of $(\mathbb Z_{10})^\times$ is $\{[1],[3],[7],[9]\}$, and
you could make the isomorphism from  $(\mathbb Z_{10})^\times$ to $\mathbb Z_4$ explicit
by mapping $[1]\mapsto0$, $[3]\mapsto1$, $[9]\mapsto2$, and $[7]\mapsto3$, for example.
A: The order of $\Bbb Z_n^\times$ is $\varphi(n)$, where $\varphi$ is the totient function.  $\varphi(10)=4$.
You can check that $3$ has order four in $\Bbb Z_{10}^\times$.  Because $3^2\equiv9, 3^3\equiv7\bmod10$.  That is, $3$ is a primitive element $\bmod10$.
Thus we have the cyclic group of order $4$.
