Counting elements of order 6 in $D_{12} \times Z_2$

Question

Actually I know how to count number of elements of particular order from direct product .But In this my counting is not matches with answer so I wanted to know where is my mistake lies .
I wanted to Count elements of order 6 in $D_{12} \times Z_2$
$6=3 \times 2=6\times 1$
There are 2 elements of order 3 in $D_{12}$ and $1$ element in $Z_2$ Similarly there are 2 elements of order 6 in $D_{12}$ and for 6 we can chose 1 or 2 as gcd remain 6 so we have 2 elements from $Z_2$ so 2*2+2=6 choices are available but answer is 4 where is my mistake lie? Any Help will be appreciated

Answer 1

Your count is correct!

To find the number of elements of order $6$ in $D_{12} \times \Bbb{Z}_2$, we have $$6=\vert (a,b) \vert =\text{lcm}\;\{\vert a \vert, \vert b \vert\}$$

so we have three cases:

Case i): $\vert a \vert =6$ and $\vert b \vert=1$

Case ii): $\vert a \vert =6$ and $\vert b \vert=2$

Case iii): $\vert a \vert =3$ and $\vert b \vert=2$

By this result " If $d$ is a positive integer, $d \neq 2$ and $d \vert n$, then the number of elements of order $d$ in $D_n$ is $\phi(d)$", each cases have two choices and hence totally we have six candidates!