1

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.

Nex
  • 3,852
idm
  • 11,824
  • 1
    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. – Noah Olander Dec 19 '15 at 14:23
  • Possible duplicate of Generators of Symmetric and Alternating Group – Dietrich Burde 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. – Dietrich Burde Dec 19 '15 at 14:30
  • Notice that in my question, $p$ is a prime number. – idm Dec 19 '15 at 14:31
  • idm, defining a group by generators and relations means that only the relations that follow from the given ones (and group axioms) should hold. If you look at $S_p$, $\sigma=(123\cdots p)$, $\tau=(12)$, then there will be other relations as well. For example $(12)(123\cdots p)=(23\cdots p)$ has order $p-1$. But $(\tau\sigma)^{p-1}=1$ is not a consequence of relations $\tau^2=1=\sigma^p$. – Jyrki Lahtonen Dec 19 '15 at 14:35
  • Defining groups by generators and relations is tricky business. For example no method exists for telling whether two elements of the group, given in terms of generators and relations, are equal. – Jyrki Lahtonen Dec 19 '15 at 14:39
  • Ok, I see. So is there a notation to says that it's generated by $\tau$ and $\sigma $ ? Maybe only $S_p=<\tau,\sigma >$ where $\tau$ is a transposition and $\sigma $ a cycle ? – idm Dec 19 '15 at 14:39
  • 1
    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 $\mathbf S_n$. – Myself Dec 19 '15 at 15:32
  • Is it correct to write $S_p=<t,s>$ where $t$ is a transposition and $s$ a $p-$cycle ? I often see this notation for this, but it is really correct ? – user301068 Dec 30 '15 at 19:30

1 Answers1

5

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 $f:G\to S_p, \sigma\mapsto (123\cdots p), \tau\mapsto (12)$.

Jyrki Lahtonen
  • 133,153