Someone told me the only transitive subgroup of $A_6$ that contains a 3-cycle and a 5-cycle is $A_6$ itself.
(1) What does it mean to be a "transitive subgroup?" I know that a transitive group action is one where if you have a group $G$ and a set $X$, you can get from any element in $X$ to any other element in $X$ by multiplying it by an element in $G$. Is a transitive subgroup just any group that acts transitively on a set? And if so, does its transitiveness depend on the set it's acting on?
(2) Why is $A_6$ the only transitive subgroup of $A_6$ that contains a 3-cycle and a 5-cycle?
Thank you for your help 🙂
Best Answer
A subgroup of the symmetric group $S_n$ is said to be "transitive" if it acts transitively on the set $\{1,2,\ldots,n\}$ via the natural action (induced by $S_n$).
So here, you would be looking at a subgroup of $A_6$ that acts transitively on $\{1,2,3,4,5,6\}$, in the natural way (i.e., via the permutation action).
Yes, the transitivity of an action depends on the set being acted on. $S_n$ acts transitively on $\{1,2,\ldots,n\}$ via its natural action; but $S_n$ also acts on $\{1,2,\ldots,n\}\times\{n+1,n+2,\ldots,2n\}$ (by acting on $\{1,2,\ldots,n\}$ via the natural action, and acting on $\{n+1,\ldots,2n\}$ by letting $\sigma$ map $n+i$ to $n+\sigma(i)$). This action is not transitive, since $1$ is never mapped to $n+1$. But in the case of groups that have "natural actions", one usually speaks of "transitive", the action being understood.
Suppose $H\lt A_6$ is transitive, and without loss of generality that it contains $(1,2,3,4,5)$ (it does, up to an automorphism of $A_6$). If the $3$-cycle in $H$ fixes $6$, then the $3$-cycle and the $5$-cycle generate the copy of $A_5$ inside of $A_6$. Conjugating by appropriate elements of $H$ you get copies of $A_5$ fixing each of $1$, $2,\ldots,6$ sitting inside of $H$, so $H$ contains all $3$-cycles, hence $H=A_6$.
If the $3$-cycle does not fix $6$, say the $3$-cycle is $(i,j,6)$; there is an element $h\in H$ that maps $6$ to $1$. If $h$ does not map $i$ nor $j$ to $6$, then conjugating $(i,j,6)$ by $h$ drops us to the previous case. If $h$ does map $i$ or $j$ to $6$, then conjugating the $3$-cycle by an appropriate power of $(1,2,3,4,5)$ gives us a $3$-cycle $(i',j',6)$ such that $h(i',j',6)h^{-1}$ fixes $6$, and we are back in the previous case again.