Finding the Left and Right Cosets in $A_4$ I'm really struggling with a Group theory class and would love some help. HW Question is as follows.

Consider the subgroups $H = \left<(123)\right>$ and $K=\left<(12),(34)\right>$ of the alternating group $G=A_4$. Carry out the following steps for both subgroups.
a.) Write $G$ as a disjoint union of the subgroup's left cosets.
b.) Write $G$ as a disjoint union of the subgroup's right cosets.
c.) Determine whether the subgroup is normal in $G$.

I honestly don't know where to start. Any help would be greatly appreciated.
 A: I'll do the first one for you.
$H=<(123)>=\{(1),(123),(132)\}$. A left coset of $H$ is a set of the form $aH$, where $a\in A_{4}$. One thing you probably learned is that two left cosets are equal to one another or disjoint. Note that
$$
\begin{aligned}
&(12)(34)H=\{(12)(34), (243),(143)\}\\
&(13)(24)H=\{(13)(24), (142), (234)\}\\
&(14)(23)H=\{(14)(23), (134), (124)\}.
\end{aligned}
$$
Thus, 
$$
A_{4}=H\sqcup(12)(34)H\sqcup(13)(24)H\sqcup(14)(23)H,
$$
where $\sqcup$ denotes a disjoint union.
Similarly, for right cosets, we have
\begin{aligned}
&H(12)(34)=\{(12)(34), (134),(234)\}\\
&H(13)(24)=\{(13)(24), (243), (124)\}\\
&H(14)(23)=\{(14)(23), (142), (143)\}.
\end{aligned}
Thus,
$$
A_{4}=H\sqcup H(12)(34)\sqcup H(13)(24)\sqcup H(14)(23).
$$
Now, note that $(12)(34)H\not=H(12)(34)$. Thus, $H$ is not normal in $A_{4}$.
Recall that when computing cosets, the representative is not unique. That is, I used $(12)(34),(13)(24), (14)(23)$ and $(1)$ as the coset representatives. The idea is you start with $H$. Then, to get a different coset, choose $a\in A_{4}\setminus H$ to get $aH$. Then, to get another coset, choose $b\in A_{4}\setminus(H\cup aH)$, and so on, until you have all of the elements in $A_{4}$ contained in one of the cosets.
A: I'll just explain the general process, using $H = \langle (123) \rangle = \{1, (123), (132)\}$.
Remember that by definition, the set of left cosets of $H$ is the set of sets $\{gH : g \in A_4\}$, where each coset $gH = \{gh : h \in H\}$ is a set of group elements.
First, verify that $1H = (123)H = (132)H = H$, since $H$ is a subgroup hence closed under multiplication. You can see the computation of the next coset and then circle back to this spot, if you're not sure yet how to verify this.
We get a new coset $gH$ by choosing some group element $g \not\in H$. For example, we can find $(13)(24)H$ by (left) multiplying everything in $H$ by $(13)(24)$,
\begin{array}{|c|c|r l|} \hline
g & h \in H & gh \in gH& \\ \hline
(13)(24) & 1 & (13)(24) \cdot 1 &= (13)(24) \\\hline
(13)(24) & (123) & (13)(24) \cdot (123) &= (142) \\\hline
(13)(24) & (132) & (13)(24) \cdot (132) &= (234) \\\hline
\end{array}
and so we see $(13)(24)H = \{(13)(24), (142), (234)\}$.
To keep finding more left cosets of $H$, pick some $g$ that hasn't shown up in any coset so far and compute $gH$ (so don't choose $g$ in $H$ or $(13)(24)H$, for example). This is because cosets are either equal, or disjoint. You should even take the time to verify an example of this, if it sounds at all unlikely: Compute $(142)H$ and verify that it's the exact same set as the $(13)(24)H$ we just computed.
You can continue in this manner to write $H$ as a union of disjoint left cosets. For example, next you could choose $g = (12)(34)$, as it hasn't shown up in the cosets $H$ or $(13)(24)H$. You'd get the coset $(12)(34)H$ and have only one remaining, after that.
To find right cosets of $H$, it's essentially the same process. For example, to compute $H(13)(24)$,
\begin{array}{|c|c|r l|} \hline
h \in H & g & hg \in Hg& \\ \hline
1 & (13)(24) & 1 \cdot (13)(24)         &= (13)(24) \\\hline
(123) & (13)(24) & (123) \cdot (13)(24) &= (243) \\\hline
(132) & (13)(24) & (132) \cdot (13)(24) &= (124) \\\hline
\end{array}
so that $H(13)(24) = \{(13)(24), (124), (243)\} \neq \{(13)(24), (142), (234)\} = (13)(24)H$. Already we can see that $H$ can't be normal in $G$, as we've exhibited some $g \in A_4$ for which $gH \neq Hg$.
