# Show that $S_p=\langle \tau,\sigma \mid \tau^2,\sigma ^p\rangle$

The fact that $$\langle\tau,\sigma \mid \tau^2,\sigma ^p\rangle\ \subset\ S_p$$ is obvious. But how can I show the other inclusion? $p$ is a prime number. The initial question is to show that $S_p$ is generated by a transposition and a $p-$cycle. It looks to be different than $\langle\tau,\sigma \mid \tau^2,\sigma ^p\rangle$ but I don't understand why.

• The group $\langle \sigma, \tau | \tau^2, \sigma^p\rangle = \mathbb{Z}_2*\mathbb{Z}_p$ is infinite, so that's not true. Maybe you actually want to show that $S_p$ is a homomorphic image of the other group? This follows since $(1 2)$ and $(1 2 \cdots p)$ generate $S_p$ and application of the universal property of free groups. Dec 19 '15 at 14:23
• Possible duplicate of Generators of Symmetric and Alternating Group Dec 19 '15 at 14:25
• @NoahOlander: Yes, I want to prove that $S_p$ is generated by a transposition and a $p-$cycle. But it doesn't meant that $S_p=<\tau,\sigma\mid \tau^2,\sigma ^p>$ ? I don't understand... it look to be exactly the same thing to me. If not, why ? And how to show that $S_p$ is generated by a transposition and a p-cycle ?
– idm
Dec 19 '15 at 14:29
• @idm See the duplicates on MSE how to generate $S_p$ by an $n$-cycle and a $2$-cycle. Dec 19 '15 at 14:30
• To answer your original question, if $G\subseteq \mathbf S_n$ contains $\sigma = (1\ 2\ \dots\ n)$ and $\tau = (1\ 2)$, then it contains $\tau^\sigma = (2\ 3)$, $\tau^{\sigma^2} = (3\ 4)$ etc all the way to $(n-1\ n)$. From there, since $(a\ b)^{(b\ c)}=(a\ c)$ its easy to see that $G$ contains all transpositions and then $G$ must be $\mathbf S_n$. Dec 19 '15 at 15:32

This is false as stated. The group $$G=\langle \tau,\sigma\mid \tau^2=1=\sigma^p\rangle$$ is always infinite (if $p>1$). It does have $S_p$ as its quotient. There is a surjective homomorphism $f:G\to S_p, \sigma\mapsto (123\cdots p), \tau\mapsto (12)$.