Powers of Permutations : All possible products generated by $(12),(13)$ Let $e$ denote the identity permutation, $\sigma = (12)$, and $\tau = (13)$.
I understand that $\langle \sigma \rangle$ would be taking powers of $\sigma$. So $\langle \sigma \rangle$ would give me $(12), (12)(12) = (1)(2) = e.$
Now I want to consider $\langle (12),(13) \rangle$. From above $(12)^{2} = e$, and $(13)^{2} = e$. Then $(12)(13) = (132).$ Also $(12)(12)(13) = (12)(132) = (13).$
My question is I could never get $(13)(12) = (123)$ right? And have I got all possible products generated by $(12),(13)$?
Thanks in advance.
 A: $\langle A\rangle$ must contain every permutation that is a product of any number of members of $A$, in any order, with repetitions allowed.
Thus, for starters $\langle(12),(13)\rangle$ must include every power of $(12)$ and every power of $(13)$; that gets you $e, (12)$, and $(13)$. It must also contain $(12)(13)=(132)$ and $(13)(12)=(123)$ and every product of either of these with the other or with $(12)$ or $(13)$. The only member of $S_3$ that we’re missing is $(23)$, and it is in $\langle(12),(13)\rangle$; can you see how to get it as a product of two of the members that we already have?
A: By definition, $\langle \sigma,\tau\rangle$ is the smallest subgroup that contains both $\sigma$ and $\tau$, and this is equal to all possible products of finite length that use $\sigma$, $\tau$, $\sigma^{-1}$, and $\tau^{-1}$, in any order. 
So you will certainly also include $\tau\sigma$; and you will also have $\sigma\tau\sigma\tau = (132)(132) = (123)$
Perhaps you are confusing $\langle \sigma,\tau\rangle$ with the set $\langle\sigma\rangle\langle\tau\rangle$, which consists of all products $\phi\psi$, where $\phi\in\langle \sigma\rangle = \{e,\sigma\}$ and $\psi\in\langle \tau\rangle = \{e,\tau\}$. Note that in general, if $H$ and $K$ are subgroups of $G$, then 
$$HK = \{hk\mid h\in H,k\in K\}.$$
We always have $HK\subseteq \langle H,K\rangle$, but $HK$ is not always a subgroup, so you don't always have equality. If $HK$ is a subgroup, then you do have equality; but here the product is not a subgroup.
A: The moment you got four different permutation in $\langle (1 2), (1 3)\rangle$ you should know that you must get $S_3$.
Your group contains $e, (1 2), (1 3)$ and $(1 2 3)$. Thus it is a subgroup of $S_3$ with at least 4 elements. Its order the must divide $6$, which is the order of $S_3$.
Since the order of $\langle(1 2), (1 3)\rangle$ is a divisor of 6, which is at least $4$, it has to be 6.
A: It's generally easier to compute powers if you write your permutations as products of disjoint cycles.  For example, $\sigma \tau = (12)(13)$ is better expressed as the single cycle $(132)$.  Let's call this element $\alpha$, for short, so $\alpha = (132)$.  You need to compute $\alpha^2 = (132)(132)$, $\alpha^3$, etc.
Please note that $(12)(12)(13)$ is not a power of $\alpha$; instead $[(12)(13)]^2 = (12)(13)(12)(13)$.
EDIT: If you want to compute $\langle (12), (13) \rangle$, you have to consider all possible products of these two elements.  Since $(12)^2 = (13)^2 = e$, it suffices to consider alternating products of $(12)$ and $(13)$, such as $(12)(13)$ and $(12)(13)(12)(13)(12)$.  It turns out you can get every element of $S_3$ in this way, which you'll hopefully discover if you experiment enough with the possible combinations.  Hint: You only need to consider products of $\leq 3$ elements.
