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Can someone tell me how I can show the following:

Consider the symmetric group $S_5$ and show that $(12345)$ and $(12)$ generate the group.

Glorfindel
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lerx
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  • What does "creates" mean here? If it means that those two elements together form a group, then it is false. If it means "generates" then this is trivial by the definition of the word. – Tobias Kildetoft Dec 13 '18 at 09:12
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    @TobiasKildetoft I think the only interesting variation on this is "show that the two elements generate all of $S_5$". So a mix-up between "a group" and "the group". – Arthur Dec 13 '18 at 09:13
  • @Arthur Ahh, of course, that makes much more sense. – Tobias Kildetoft Dec 13 '18 at 09:14
  • Thanks for your replies. Sorry @all, I meant "the" group, not a group. – lerx Dec 13 '18 at 09:24

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So, thats everything I have to show? Or is there something more to do?

Let $c = (1, 2, \dotsc, 5)$. We see that \begin{align*} c (1, 2) c^{-1} &= (2, 3) \\ c (2, 3) c^{-1} &= (3, 4) \\ &\vdots \\ c (5-2, 5-1) c^{-1} &= (5-1, 5), \end{align*} so that $(i, i+1) \in \langle (1, 2), c \rangle$ for all $1 \leq i \leq 5-1$. Next, we have \begin{align*} (2, 3) (1, 2) (2, 3)^{-1} &= (1, 3) \\ (3, 4) (1, 3) (3, 4)^{-1} &= (1, 4) \\ &\vdots \\ (5-1, 5) (1, 5-1) (5-1, 5)^{-1} &= (1, 5), \end{align*} so that $(1, i) \in \langle (1, 2), c \rangle$ for all $1 \leq i \leq 5$. Choose any $1 \leq i < j \leq 5$, then $$ (i, j) = (1, i) (1, j) (1, i)^{-1} \in \langle (1, 2), c \rangle. $$ Therefore, $\langle (1, 2), c \rangle$ contains all transpositions. Hence, $\langle (1, 2), c \rangle = S_5$.

lerx
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