Clever way to find cosets $H = \{ (1), (12)(34), (13)(24),(14)(23)\}$ an $H$ is a subgroup of $A_4$.  I know that $H$ has 3 cosets in $A_4$ by Lagrange's Theorem, but is there a clever way to compute them without going through every element in $A_4$?
 A: The left cosets are given by $\{gH:g \in G\}$. 
Notice that $a \in bH \iff b \in aH$.
If $a \in bH$, then there exists a $h \in H$ such that $a=bh$. If $a=bh$ then $ah^{-1} = b$. Since $H<G$ is a subgroup, $h^{-1} \in H$ and so $b \in aH$. The converse is also true.
What good is this? 
Well, you might think that you need to do $gH$ for each an every $g \in G$. But once an element turns up in a coset, you can cross it off your list. 
In your case, you can cross $(1)$, $(12)(34)$, $(13)(24)$ and $(14)(23)$ off your list. There's no point finding $(1)H$, $(12)(34)H$, $(13)(24)H$ or $(14)(23)H$; you won't get a new coset, you'll just get $H$ again.
Now pick an element that hasn't be crossed off, say $(132)$.
$(132)H = \{(132), (234), (124), (143) \}$, so we have another coset and can cross off $(132)$, $(234)$, $(124)$ and $(143)$. There's no point finding $(132)H$, $(234)H$, $(124)H$ or $(143)H$; you won't get a new coset, you'll just get $(132)H$ again.
Continue like this.
In general, since $|H|$ divides $|G|$, and all cosets have the same number of elements, you'll have to do this $|G|/|H|$ times to get all of the cosets.
As @Bungo says in the comments: once you have the penultimate coset, you get the last one for free: it'll be whatever's left.
