# 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 $(12),(13)$?

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Do you mean $\langle(12)(13)\rangle$, as you wrote it, or $\langle(12),(13)\rangle$, as you seem to have computed it? $\langle(12)(13)\rangle$ is simply the powers of the permutation $(12)(13)=(132)$, which are $e,(132)$, and $(132)^2=(123)$. $\langle(12),(13)\rangle$ is everything that you can get by taking products of powers of $(12)$ and $(13)$ in any order. – Brian M. Scott Nov 16 '11 at 20:42
@Brian: I meant $\langle (12), (13) \rangle$ – Student Nov 16 '11 at 20:45
@Brian: Let me see if I understand what you are saying. Does $\langle (12), (13) \rangle = \langle (13), (12) \rangle = S_{3}$? – Student Nov 16 '11 at 20:47
Yes, exactly so. – Brian M. Scott Nov 16 '11 at 20:51

$\langle A\rangle$ 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?

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I think I see it: $(12)(123) = (23)$ – Student Nov 16 '11 at 20:54
@Jon: Looks good! – Brian M. Scott Nov 16 '11 at 20:58

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.

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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.

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It's better to use \langle and \rangle, which are delimiters, instead of < and >, which are operators. – Arturo Magidin Nov 16 '11 at 21:14

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.

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