Order of elements in $D_4 \times S_3$. This problem will deal with the group $G = D_4 \times S_3$.
How many elements of each order do the groups have $D_4$ and $S_3$ have?  Using this info, determine how
many elements of each order $G$ has.  (Of course, it would be extremely tedious to compute all 48
orders one by one. You should NOT be doing that. Perhaps organize your work in a table.)
I can't figure out a good way to do this. Any help would be appreciated, thank you.
EDIT: To find the subgroups of G which are isomorphic to $Z_2 \times Z_2$, do I have to find all the elements with an order of 4, which are non-abelian?
 A: Hint: if $x \in G, y \in H$, then in $G \times H$, the $ord((x,y))=lcm(ord(x),ord(y))$
A: Elements of order $2$ will come from produc of an element of order $\leq 2$ in $D_4$ and an element of order $\leq 2$ in $S_3$ (not both taken identity). Count them.
Elements of order $4$ will be product of elements of order $4$ inside $D_4$, with elements of order $\leq 2$ in $S_3$ (count them).
Elements of order $3$ lie in Sylow-3 subgroup, which is unique here, inside $S_3$, count them.
Elements of order $6$ will come from product of an element of order $3$ and commuting element of order $2$. The elements of order $2$ commuting with element of order $3$ lie in $D_4$, count how many you get.
Sylow-2 subgroup has exponent 4, so no element of order 8.
Elements of order 12 will come from element of order $3$ and commuting element of order $4$. Elements of order $3$ are in $S_3$ and with them commuting elements of order $4$ are in $D_4$, count how many you get.
The maximum order of any element in $D_4\times S_3$ is product of maximum order of element of $D_4$ (which is 4) and of $S_3$ (which is 3). So we covered all elements (except identity :P)
