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.
Best Answer
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.