0
$\begingroup$

In my notes there is the following: $$S_3=\{ id, (1 \ 2), (1 \ 3), (2 \ 3), (1 \ 2\ 3), (1\ 3\ 2)\}$$

The subgroups are: $id, S_3, \langle (1 \ 2)\rangle, \langle (1 \ 3)\rangle, \langle (2 \ 3)\rangle, A_3=\langle (1 \ 2 \ 3)\rangle=\{id, (1\ 2\ 3), (1\ 3\ 2)\}$.

Why are all the subgroups generated by one element? We have that $|S_3|=3!=6$. So the subgroups can have the orders $1,2,3,6$. The ones with order $2$ are cyclic, and so they are generated by one element, right? Why aren't there subgroups of order $3$ ?

Also why are only the $S_3$ and $A_3$ trasitive?

  • 3
    The subgroup $A_3$ has order $3$. – 2017-01-06
  • 1
    If a group $G$ is acting on a set of $n$ objects, it's "difficult" for the group to act transitively if $|G| < n$; the orbit of an object can only be as big as $|G|$. – 2017-01-06
  • 0
    A group action is transitive if starting from any point we can "get to" any other point via the action of $G$. Why is this not possible for example for $\langle (1 \ 2)\rangle$ ? @pjs36 – 2017-01-06
  • 1
    In this case, what's the orbit of $3$? (I assume our subgroups are acting on $X = \{1,2,3\}$ in the usual way) More generally, if $G$ is acting on $X$, the orbit of $x \in X$ is the set $\operatorname{Orb}_G(x) = \{g \cdot x : g \in G\}$ (can you see why $\lvert\operatorname{Orb}_G(x)\rvert \le |G|$?). Recall that this action is transitive if and only if $\operatorname{Orb}_G(x) = X$. So based on cardinality, any subgroup of order less than $3$ can't act transitively on $\{1,2,3\}$ here. – 2017-01-06
  • 0
    I got it!! :-) @pjs36 – 2017-01-06
  • 1
    @MaryStar Why have have you deleted your question ? I was just writing an answer. https://math.stackexchange.com/questions/2093182/height-normal-distribution – 2017-01-11
  • 0
    Sorry. I opened it again. @callculus – 2017-01-11
  • 1
    @MaryStar You was on the right track. I posted an answer. – 2017-01-11

2 Answers 2

3

Hint: Recall that every group of prime order is cyclic and by Lagrange's theorem the only possible order for non trivial subgroups of $S_3$ are $2$ and by $3$ which are of course prime. ($S_3$ is not cyclic,not even abelian)

  • 0
    I understand!! I have also an other question... We have that $\sigma=k_1k_2\ldots $, where $k_1, k_2, \ldots$ are disjoint cycles of $S_p$, where $p$ a prime. Then we have the following: $$p=ord(\sigma)=lcm \ (ord(k_i)), \ i=1, 2, \ldots$$ Why does this imply that $ord(k_1)=p$ ? – 2017-01-06
  • 0
    @MaryStar: Since the question is completely different,i'd suggest you to ask this as a separate question! – 2017-01-08
1

But there's a subgroup of order $3$ in your list , viz. $A_3$. Being a group of prime order it has to be cyclic. And since it includes both elements of order $3$ from the original group, this is the only subgroup of order $3$ that we can possibly generate here. In other words, $\langle(1\,2\,3)\rangle=\langle(1\,3\,2)\rangle=A_3$, and there are no other elements of order $3$.

  • 0
    I understand!! I have also an other question... We have that $\sigma=k_1k_2\ldots $, where $k_1, k_2, \ldots$ are disjoint cycles of $S_p$, where $p$ a prime. Then we have the following: $$p=ord(\sigma)=lcm \ (ord(k_i)), \ i=1, 2, \ldots$$ Why does this imply that $ord(k_1)=p$ ? – 2017-01-06