Let $G = \{1, 8, 14, 18, 47, 51, 57, 64\}$. I would like to find all elements of $G$ with order $4$. I know that $G \cong \mathbb Z_4 \oplus \mathbb Z_2$, and that the elements of $\mathbb Z_4 \oplus \mathbb Z_2$ of order 4 are $(1,0)$, $(1,1)$, $(3,0)$, and $(3,1)$.
But an answer text says that since $8^2 = 64 \neq 1$, $8$ has order $4$ in G. My question is how did they get this? Is there a shortcut to quickly calculating that order?
Also, is there a way of calculating $64(8)$ and $14^2$ quickly as well? And how does the fact that 14 $\notin <8>$ deduce that the elements of order $4$ in $G$ are $8$, $57$, $8*14$, and $8*57$?
Thanks in advance