Ok, so I know how to do this by direct computation, but that seems like it will take a long time, and that is probably not how my professor wants this done. This is a practice question for the exam, and I am worried that if I run into this question on the exam, I will not be able to do this except for through direct computation. $K$ in this question is the Klein 4 group $\{I, (12)(34), (13)(24), (14)(24)\}$, $A_3$ is the alternating group $\{I, (123), (321)\}$ and $D_4$ is the dihedral group.
I feel like I need to do something with the conjugacy classes or cosets of the subgroups, but I'm not sure. To be honest, I'm not even sure where to start on this problem other than doing direct computations of $S_4$ in both cases. Please walk me through how you would go about solving this, and if you have any problems that are similar, I would love to hear them (his exams are always VERY similar to his practice exams, so I really want to understand this). Any help is greatly appreciated, thanks
In both cases, you could use the fact that if $H$ and $K$ are finite subgroups of a group $G$, then $|HK|=|H|~|K| / |H \cap K|$.
To show $S_4=A_3 D_4$, observe that the nonidentity elements of $A_3$ have order 3, whereas the symmetries of a square $D_4$ is a group of order 8, and 3 does not divide 8, whence $A_3 \cap D_4=1$. Thus, $|A_3 D_4|=3 \times 8 /1 = 24$, whence $A_3 D_4$ (which is a subgroup of $S_4$) exhausts all the elements of $S_4$.
To show $A_4=A_3 K$, use the fact that an even permutation $a \in A_3$ composed with another even permutation $k \in K$ is even, whence $A_3 K \subseteq A_4$. As before, $A_3 K$ has order $3 \times 4=12$, which is the same as $|A_4|$. This implies the two sets $A_3 K$ and $A_4$ are equal.