Question:
Without doing any calculations in $Aut\left ( \mathbb{Z}_{20} \right )$, determine how many elements of $Aut\left ( \mathbb{Z}_{20} \right )$ have order 4. How many have order 2?
Attempt:
By a certain theorem: $Aut\left ( \mathbb{Z}_{20} \right )\cong U\left ( 20 \right )$. Recall again that by the property of isomorphism from a group G to a group $\bar{G}$: the number of elements of a certain order in G equals the number of element of the same order in $\bar{G}$.
The order of $U\left ( 20 \right )$ is 8.
Recall: For a group of order n, the number of elements of order divisor d of n is represented by the phi-euler totient $\phi\left ( d \right )$.
From this, it follows that there are 2 elements of order 4 and 1 element of order 2 in $U\left ( 20 \right )$. By the property of isomorphism: the number of elements of such orders are preserved in $Aut\left ( \mathbb{Z}_{20} \right )$.
Is this correct? Thanks in advance.
First note that:
$Aut\left ( \mathbb{Z}_{20} \right ) \cong U\left ( 20 \right )\cong U(4)\oplus U(5) \cong \mathbb{Z}_2 \oplus \mathbb{Z}_4$
For finding the number of element having order 4, suppose $(x,y) \in \mathbb{Z}_2 \oplus \mathbb{Z}_4$. Since $|(x,y)| = lcm(|x|,|y|)$, $x$ is free and $y$ is of order 4. Since there are $\phi(4) = 2$ elements in $\mathbb{Z}_4$ have order 4, there must be 4 elements are of order 4 in $\mathbb{Z}_2 \oplus \mathbb{Z}_4$, so is $Aut\left ( \mathbb{Z}_{20} \right )$.
For finding the number of element of order 2, sicne $4\cdot (x,y) = 0$ for every pair of $(x,y)$, by Lagrange theorem, the possbile order of elements are 1, 2, 4. You have calculated that $\phi(20) = 8$, so among these 8 elements, 4 of them are of order 4, and only identity is of order 1, so this implies there are 3 elements which are of order 2.