In the alternating group $A_4$ for the subgroup of $H\le A_4$ find all of its left and right cosets where $$H=\{\text{Id},(1,2,3),(1,3,2)\}.$$
We have by Lagrange's theorem that $$|A_4:H|=\frac{|A_4|}{|H|}=\frac{24/2}{3}=4$$ which means that $H$ has $4$ left and $4$ right cosets.
By definition a left coset of $H$ in $A_4$ determined by $x\in A_4$ is the set $\{xh:h\in H\}$.
So, the left cosets will be $x\text{Id},x(1,2,3),x(1,3,2)$, right?
How can I find the $4$th coset?
On
First note that $$A_4=\{\text{Id},(1,2,3),(1,3,2),(2,4,3),(2,3,4),(1,2)(3,4),(1,2,4),(1,3,4),(1,3)(2,4),(1,4,2),(1,4,3),(1,4)(2,3)\}$$ We have, \begin{align} (1,2)(3,4)\cdot(1,2,3)=(2,4,3)\\ (1,3)(2,4)\cdot(1,2,3)=(1,4,2)\\ (1,4)(2,3)\cdot(1,2,3)=(1,3,4) \end{align} and \begin{align} (1,2)(3,4)\cdot(1,3,2)=(1,4,3)\\ (1,3)(2,4)\cdot(1,3,2)=(2,3,4)\\ (1,4)(2,3)\cdot(1,3,2)=(1,2,4) \end{align} So the left cosets of $H$ are $\text{Id}H,(1,2)(3,4)H,(1,3)(2,4)H,(1,4)(2,3)H$.
We have to repeat the above calculations for $Hx$ (right coset).
Obviously, the 100% way to do this is get every element of $A_4$ multiply it to left and right of $H$ to get your $aH$ and $Hb$. I will answer to answer how to come up with the selection of elements to get 4 left cosets and 4 right cosets.
I am copying the listing of $A_4$ by johnny09 posted in his or her answer, but with a rearrangement for emphasizing certain elements: $$A_4=\{\text{Id},$$
$$(1,2,3),(1,3,2),(2,3,4),(2,4,3),(1,2,4),(1,4,2),(1,3,4),(1,4,3)$$
$$(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\}$$
Looking at $A_4$ this way, we have 3 to cosets derive because $H$ is its own coset, and we also have 3 elements other than the identity that are not like the others (We can also say we have 4 cosets to derive and 4 elements that are not like the others). I am going to base my selection of elements on these 3 elements.
A permutation property is: $$(a,b)(c,d)=(adc)(abc)$$
From this, we can write:
$$(1,2)(3,4) = (143)(123) \tag{1}$$
$$(1,3)(2,4) = (142)(132) \tag{2}$$
$$(1,4)(2,3) = (132)(142) \tag{3}$$
A permutation property is: $$(a,b)(c,d)=(dba)(dca)$$
From this, we can write:
$$(1,2)(3,4) = (421)(431) \tag{4}$$
$$(1,3)(2,4) = (431)(421) \tag{5}$$
$$(1,4)(2,3) = (341)(321) \tag{6}$$
My selection for $a$ to derive left cosets $aH$ will be based on $(1),(2),(6)$
$$a = 1, (143), (142), (341)$$
My selection for $b$ to derive right cosets $Hb$ is the reverse of $(1),(2),(6)$:
$$(1,2)(3,4) = (143)(123) \tag{1}$$
$$(1,3)(2,4) = (142)(132) \tag{2}$$
$$(1,4)(2,3) = (341)(321) \tag{6}$$
$$(1,2)(3,4) = (321)(341) \tag{1(rev)}$$
$$(1,3)(2,4) = (231)(241) \tag{2(rev)}$$
$$(1,4)(2,3) = (123)(143) \tag{6(rev)}$$
I select $$b = 1, (341), (241), (143)$$
To prove that my choices for a work, I used the property: $(a,b,c)(c,b,e)=(abe)$. Other permutation properties are: $(a,b)(c,d)=(cda)(cba)=(acd)(abd)=(dab)(dcb)$. For b, see: Why did this reversal from the left cosets of $\langle (1, 2, 3) \rangle$ in $A_4$ give me the right cosets?
For a systematic approach (but still aiming for the products of disjoint transposition), we could try to solve $(1,2)(3,4)=(123)(abc)$, $(1,2)(3,4)=(132)(abc)$ and also for the others.