Can someone please help me with the last part of this problem I am working on? All my work is below. Thank you for your time and help.
Let $N = 10m,$ where $m> 1$ and $5\nmid m.$
i. Let $G_1$ be a group of order $N$ containing $4m$ elements of order 5, and no element of order 10. How do the elements of order 5 divide into conjugacy classes.
$\textit{Proof.}$ Given that the Sylow 5-subgroups are of order 5, we can let $P$ be one of them. Here, the number of Sylow 5-subgroups is $m.$ It follows that $|G: N_G(P)| = m$ and then $|N_G(P)| = 10.$ Since the group $G$ has no elements of order 10, then $C_G(P) = P.$ Thus, $N_G(P)$ contains two classes of conjugate elements of order 5. Then, it follows that $G$ also contains exactly two classes of conjugate elements of order 5. Notice that we use the fact that all Sylow 5-subgroups of $G$ are conjugate to one another.
ii. Let $G_2$ be a group of order $2N$ that contains $4m$ elements of order 5, and no element of order 10. How do the the elements of order 5 divide into conjugacy classes.
$\textit{Proof.}$ By the same process in part i. we see that in this case, we get exactly one conjugacy class of elements of order 5.
iii. How do I provide an example of a group $G_1$ and a group $G_2?$
$\textit{My idea.}$ Can we apply $D_{10}$ to work here for $G_1$? I am not sure about $G_2.$