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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 :)

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$S_n$ and $A_n$ have "canonical" actions since they are defined as groups of permutations. They act on any set of "n" elements by permutation. – Vhailor May 20 '11 at 3:04
Oh I see. Thanks. – badatmath May 20 '11 at 3:12
So if $S_n$ is the Galois group of some polynomial, the elements in the cycles are just the roots of that polynomial? Like (1 2 3) would send the first root to the second root, etc? – badatmath May 20 '11 at 3:16
I have no idea what is going on, though, I would like to have someone put all these clear; just peut-être; can someone do this? Thank you very much. – awllower May 20 '11 at 3:39
@badatmath: The Galois group of a polynomial is a group of automorphisms of some field. Formally, it is not equal to $S_n$. However, the action of the Galois group on the roots induces a homomorphism into a symmetric group, because whenever a group $G$ acts on a set $X$, this induces a homomorphism $G\to S_X$. Saying that the Galois group of "some polyomial" is $S_n$ doesn't really tell the whole story. If the polynomial is irreducible of degree $n$, and the Galois group is isomorphic to $S_n$, then this isomorphism can be realized by looking at the action on the roots. (cont) – Arturo Magidin May 20 '11 at 3:43
up vote 8 down vote accepted

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.

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thanks for your help! to prove that the 3-cycle and 5-cycle generate $A_5$, do you use a similar argument and induct downwards? – badatmath May 20 '11 at 6:39
@badatmath: A subgroup of $A_5$ that contains a 3-cycle and a 5-cycle must have order a multiple of 15; cannot be 15 (all groups of order 15 are cyclic, $A_5$ has not elements of order 15). It cannot be of order 30 (it would be of index 2 in $A_5$, hence normal, but $A_5$ is simple), and it cannot be of order $45$ (order must divide $|A_5|=60$), so it is all of $A_5$. – Arturo Magidin May 20 '11 at 16:37
oh awesome, thanks! – badatmath May 21 '11 at 2:25

Here is another way to look at it:

Let $H$ be a transitive subgroup of $A_6$ containing a 5-cycle and a 3-cycle. Then $5$ divides $|H|$, and also $6$ divides $|H|$ since $H$ acts transitively. Thus $30$ divides $|H|$, and if $H$ is not all of $A_6$, by simplicity $H$ must have order $30$ or $60$. $30$ is impossible, as such a group has a (normal) subgroup of order $15$, and every group of order $15$ is cyclic. Yet there is no element of order $15$ in $A_6$. $60$ is also impossible, as every such subgroup of $A_6$ is a copy of $A_5$, which is the point stabilizer of one of $\{1,2,3,4,5,6\}$, and thus not transitive. Thus $H$ is $A_6$.

EDIT - Here is an argument that if $H$ has order $60$ and contains a 3-cycle, then it is a point stabilizer in $A_6$. First, every subgroup of $A_6$ of order $60$ is isomorphic to $A_5$, and thus $H$ is generated by its order 3 elements. We can assume up to conjugation that the 5-cycle in $H$ is $(12345)$. Since all order 3 elements in $A_5$ are conjugate, all order 3 elements in $H$ are 3-cycles. If no 3-cycle in $H$ contains a $6$, then $H$ cannot be transitive, since it is generated by these 3-cycles. Thus, assuming $H$ is transitive, and by considering appropriate powers of $(12345)$, we can also assume $H$ contains the 3-cycle $(126)$. But then $(12345)(126)=(16)(2345)$ has order $4$, and $A_5$ has no elements of order $4$.

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Hi there! Why does $H$ acting transitively mean that $6$ must divide $|H|$? – badatmath May 21 '11 at 2:51
@badatmath: $H$ acts transitively on a set of cardinality $6$; by the orbit-stabilizer theorem, if $K\le H$ is the stabilizer of a point, then $[H:K]=6$. – user641 May 21 '11 at 7:53
Oh cool, thanks! I forgot about that theorem. – badatmath May 21 '11 at 15:19

Let $H\leq A_6$ be transitive and generated by a 3-cycle and a 5-cycle.

Let if possible, $H\neq A_6$, and let us compute $|H|$.

$|H|$ is divisible by 15, and divides 360$=|A_6|$, so it is one of $\{15,30,45,60,90,120,180\}$.

  • $|H|$ can not be $\{90,120,180\}$, otherwise we get a subgroup of $A_6$ of index less than $6$.
  • $|H|$ can not be 15, since then $A_6$ will have an element of order 15, which is not possible,
  • $|H|$ can not be 45, since a group of order 45 is abelian and so it contains an element of order 15.
  • $|H|$ can not be 30, since a group of order 30 has normal Sylow-5 subgroup, and so it will contain a subgroup of order 15, hence an element of order 15.

Hence $|H|$ should be $60$.

Now in this subgroup of order 60, Sylow-5 subgroup can not be normal, since if it is normal, then it will also be normalized by an element of order 3, giving a subgroup of order 15, hence an element of order 15.

So $H$ is a group of order $60$, which has no normal Sylow-5 subgroup; $H$ must be isomorphic to $A_5$. There are 6 Sylow-5 subgroups of $H\cong A_5$, hence 24 elements of order 5; they will be 5 cycles, hence fixing an element in $\{1,2,...,6\}$. Let $(12345)\in H$.

As $H$ is transitive subgroup, there will be an element $\sigma \in H$ such that $\sigma(6)=1$, so $\sigma (12345)\sigma^{-1}\in H$, will be a 5-cycle fixing 1; in this way all Sylow-5 subgroups of $A_6$, and hence all element of order 5 of $A_6$ will be in $H$, exceeding the size of $H$.

Hence we must have $H=A_6$.

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