2
$\begingroup$

My homework question is:

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

but I am not sure how to start with.

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 $xH=\{xh:h\in H\}$.

$\endgroup$
2
  • $\begingroup$ 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 $\langle(234)\rangle$ $\endgroup$ – Daniil Feb 28 at 10:36
  • $\begingroup$ 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 $\endgroup$ – Daniil Feb 28 at 10:43
2
$\begingroup$

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 $C_3$.

$\endgroup$
1
$\begingroup$

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 $V_4\cap H=\{e\}$.)

Similarly for the left cosets.

$\endgroup$
2
  • $\begingroup$ The last two sentences are wrong. $\endgroup$ – ancient mathematician Feb 28 at 17:47
  • $\begingroup$ You're right. Thanks for pointing it out. @ancientmathematician $\endgroup$ – Chris Custer Feb 28 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.