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Suppose we have an action $\pi$ of $S_4$ on the set $\{x,y,z\}$, satisfying $\pi_{(1 2)} (x)=y$ and $\pi_{(2 3)} (y)=z$. Find $\pi_{(3 4)} (x)$ and $|Stab(x)|$.

What can I do here? I know that the action is a homomorphism from $S_4$ to the group of symmetries of $\{x,y,z\}$, but how do I say anything about $\pi_{(3 4)} (x)$ when the given $\pi_{(1 2)}$ and $\pi_{(2 3)}$ both fix $4$ when multiplied? Also, I know that the stabilizer is a subgroup of $S_4$, so its order has to divide $24$, and initially I thought that it consists of all permutations that fix $1$, so $6$ in total, but this seems to be wrong. Thanks in advance!

JBuck
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1 Answers1

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We identify $S_3$ with the permutations of the set $\{x,y,z\}$. (For $S_4$ we use the usual set $\{1,2,3,4\}$.) Then the action $\pi$ induces a homomorphism $S_4\to S_3$ with $(12)\to(xy)$, $(23)\to(yz)$, this homomorphism is surjective, so its kernel is a normal subgroup of $S_4$ of order $4!/3!=4$. There exists exactly one such subgroup $K$, it is the subgroup with elements

  • $()$, $(12)(34)$, $(13)(24)$, $(14)(23)$,
  • so $(34)$ acts like $(12)$, $(24)$ acts like $(13)$, and $(14)$ like $(23)$.

The stabilizer of $x$ in $S_3$ is the subgroup generated by $(yz)$. A preimage of $(yz)$ is $(23)\in S_4$. The group generated by $K$ and $(23)$ has eight elements: $()$, $(23)$, $(14)$, $(14)(23)$, $(12)(34)$, $(13)(24)$, $(1243)$, $(1342)$.

dan_fulea
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