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
Best Answer
Hints:
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: $$ (a, b, c)^2 = (a^2, b^2, c^2) = (1, 1, 1) = 1 $$