Left coset and right coset of subgroup $H=\langle(234)\rangle$ in alternating group $A_4$

My homework question is:

Partition $$G=A_4$$ into left cosets of the subgroup $$H=\langle (234)\rangle$$

I know that $$A_4= \{(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)\},$$ am I right?

Also, the left coset of H is $$xH=\{xh:h\in H\}$$.

• About $A_4$ elements: You know that $|A_4|=\frac{4!}{2}=12$, so you got the right number. Concerning the elements that you have, you can check if the signature of each permutation is $-1$. And to get the left cosets of $H$, you simply compose each element of $A_4$ with all the elements of $\langle(234)\rangle$ – Daniil Feb 28 at 10:36
• And to get the left cosets of $H$, you simply compose each element of $A_4$ with all the elements of $\langle(234)\rangle$. For example for $x=Id\in A_4$ you obtain the following as left coset: $xH=\{(234),(243),Id\}$. You do the same with each element of $A_4$ and you leave only different cosets. If i remember well, to check if yo got the right number of left cosets, you divide the cardinal of $A_4$ by the cardinal of $H$ (not sure about it anymore), so you will obtain $4$ left cosets – Daniil Feb 28 at 10:43

A 3-cycle has order 3, and $$A_4$$ has order 12, so there will be four left cosets by Lagrange's theorem. Recall the Klein four group is a subgroup of $$A_4$$, and it intersects trivially with $$C_3$$ since their elements have different orders, so each element of $$K_4$$ induces a left coset of $$C_3$$.
Since $$H=\{e,(234),(243)\}$$, we get that the four right cosets are \begin{align} H, &\\ H(12)(34)&=\{(12)(34),(132),(142)\}, \\ H(13)(24)&=\{(13)(24),(143),(123)\},\\ H(14)(23)&=\{(14)(23),(124),(134)\}, \end{align}
just by multiplying through by the elements of $$V_4$$. (I know that will get me the four cosets since $$V_4\cap H=\{e\}$$.)