Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :)

share|improve this question
4  
$S_n$ 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
show 4 more comments

3 Answers

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.

share|improve this answer
    
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
add comment

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 $4$.

share|improve this answer
    
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
add comment

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 $H=A_6$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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