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:
$$
(a, b, c)^2 = (a^2, b^2, c^2) = (1, 1, 1) = 1
$$
Yes, it is true. The group $\mathbb{Z}_7^\times$ is a cyclic group of order $6$ (it is generated by $3$). And the group $\mathbb{Z}_{31}^\times$ has an element of order $6$, which is $6$.
Best Answer
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 $$ f(n+m) = 3^{n+m}=3^n\cdot3^m \pmod{10} = f(n)f(m) $$