To which group is the $\mathbb Z_{20}^*$ isomorphic? I have a question saying that to which group is $\mathbb{Z}_{20}^{*}$ is isomorphic, where $\mathbb{Z}_{20}^{*}$ is the set of the not zero divisors of $\mathbb{Z}_{20}$.
Here is what i think: $\mathbb{Z}_{20}^{*}=\{1,3,7,9,11,13,17,19\}$. It has 8 elements. Then it is isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ or $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ or $\mathbb{Z}_{8}$. But how can I decide which of them? And is this group a multiplicative group or an additive group? How can I know?
Thank you
 A: To hint your last question first, if $a,b$ are each relatively prime to $20$, then $ab$ is also relatively prime to $20$.  $a+b$ need not be relatively prime to $20$.
To hint your first question, $\mathbb{Z}_8$ has an element of order 8,  $\mathbb{Z}_2\times\mathbb{Z}_4$ has an element of order 4 but no element of order $8$, while in $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$ all elements but the identity have order 2.
A: Hints:


*

*Is $1 + 3$ in $\Bbb Z^*_{20}$? On the other hand, one can show that it's closed under multiplication.

*Is $\Bbb Z^*_{20}$ cyclic? What's the order of $3$? Answer both questions and one possibility will remain.



To elaborate on the second hint. We have:
\begin{align}
3^1 &= 3 &\equiv 3 \pmod{20} \\
3^2 &= 9 &\equiv 9 \pmod{20} \\
3^3 &= 27 &\equiv 7 \pmod{20} \\
3^4 &= 81 &\equiv 1 \pmod{20} \\
\end{align}
Therefore the order of $3$ is $4$. This eliminates the possibility $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$. With similar computations, you can see that none of the elements of $\Bbb Z^*_{20}$ generates the group. Hence, it's not cyclic. One possibility remains and it's $\Bbb Z_4 \times \Bbb Z_2$.

To show that all elements of $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$ have order $1$ or $2$, consider $(a, b, c) \in \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$. We have:
$$
(a, b, c)^n = (a^n, b^n, c^n)
$$
Since each of $a$, $b$, $c$ is a member of $\Bbb Z_2$, it has order $1$ or $2$. This forces:
$$
(a, b, c)^2 = (a^2, b^2, c^2) = (1, 1, 1) = 1
$$
A: Hint By the Chinese Remainder Theorem $Z_{20}$ is isomorphic to $Z_{4}\times Z_5$. Also by the Chinese Remainder Theorem $Z_{20}^*$ is isomorphic to $Z_{4}^*\times Z_5^*$ [If you know rings, you can use directly the CRT for rings, otherwise you can prove the second statement from the standard CRT].
Since $5$ is prime $Z_5^*$ is a cyclic group with $4$ elements. $Z_{4}^*$ has 2 elements, and there are not too many groups with 2 elements...
