##S_p## generated by any transposition and p-cycle

[Bashyboy]

Homework Statement

Let $p$ be prime. Show that $S_p = \langle \tau, \sigma$, where $\tau$ is any transposition and $\sigma$ any $p$-cycle.

Homework Equations

The Attempt at a Solution

I read somewhere that it suffices to prove this for $\sigma = (1,2,...,p)$. Intuitively, this is clear, but I want to justify this. Let $\sigma = (x_1,.x_2,...,x_p)$. Let $f : \{1,...,p\} \to \{1,...,p\}$ be defined $f(x_i) = i$ and fixes everything else. It seems that we need to show that $S_p = \langle \tau, \sigma \rangle$ if and only if $S_p = \langle f \tau, f \sigma \rangle$, which can be proven by showing $\langle f \tau , f \sigma \rangle = f \langle \tau, \sigma \rangle$, in order to justify this reduction.

Showing $\langle f \tau , f \sigma \rangle \subseteq f \langle \tau, \sigma \rangle$ is rather simple, but I am having trouble with the other inclusion. I am probably overlooking some trivial fact. I could use a hint.

 

[fresh_42]

What is $\langle f\tau,f\sigma \rangle$ after you managed to prove the result for $(1,\ldots ,p)\,$?

 

[Bashyboy]

What is $\langle f\tau,f\sigma \rangle$ after you managed to prove the result for $(1,\ldots ,p)\,$?

I was thinking that $\langle gx,gy \rangle = \langle g \langle x,y \rangle$ was a result that held for all groups. Is that not true? I should have specified that I would like to prove that it is true for all groups.

 

[fresh_42]

I was thinking that $\langle gx,gy \rangle = \langle g \langle x,y \rangle$ was a result that held for all groups. Is that not true? I should have specified that I would like to prove that it is true for all groups.

This is not what I meant. Far simpler than this. I asked basically: What are $f \tau$ and $f \sigma$, what do they look like? And then, if you managed to prove the result in the special case of transposition and $p-$cycle $(1,\ldots ,p)$, what is $\langle f\tau , f \sigma \rangle$ for a subgroup?

 

[Bashyboy]

Well, $f \tau$ would just be some transposition and $f \sigma$ would give us back $(x_1,...,x_p)$, right?

 

[fresh_42]

Well, $f \tau$ would just be some transposition

Yes. Since we have no further condition on the transposition used, we can use $\tau$ as well as $f \tau$. It makes no difference for "any".

... and $f \sigma$ would give us back $(x_1,...,x_p)$, right?

Let $\sigma = (x_1,.x_2,...,x_p)$. Let $f : \{1,...,p\} \to \{1,...,p\}$ be defined $f(x_i) = i$ and fixes everything else.

O.k., beside of the part "fixes everything else" which I don't see why you want to have this, yourself has defined $\sigma = (x_1,x_2, \ldots,x_p)$ so I assumed $f \sigma = (1, \ldots ,p)$. I thought this was the formal notation for renumbering the elements, which we actually want to do. Now assume we have proven, that $S_p$ is generated by some transposition $f \tau$ and the $p-$cycle $f \sigma = (1,\ldots ,p)$. This makes $\langle f\tau , f \sigma \rangle = S_p$. You further have said

Showing $\langle f \tau , f \sigma \rangle \subseteq f \langle \tau, \sigma \rangle$ is rather simple.

which leaves us with the situation $S_p=\langle f \tau , f \sigma \rangle \subseteq f \langle \tau, \sigma \rangle$ rather simple and I simply wanted to point you to
$$
f \langle \tau, \sigma \rangle \subseteq S_p = \langle f \tau , f \sigma \rangle
$$ is by far even simpler!

 

[Bashyboy]

Yes. Since we have no further condition on the transposition used, we can use $\tau$ as well as $f \tau$. It makes no difference for "any".O.k., beside of the part "fixes everything else" which I don't see why you want to have this, yourself has defined $\sigma = (x_1,x_2, \ldots,x_p)$ so I assumed $f \sigma = (1, \ldots ,p)$. I thought this was the formal notation for renumbering the elements, which we actually want to do. Now assume we have proven, that $S_p$ is generated by some transposition $f \tau$ and the $p-$cycle $f \sigma = (1,\ldots ,p)$. This makes $\langle f\tau , f \sigma \rangle = S_p$. You further have said

which leaves us with the situation $S_p=\langle f \tau , f \sigma \rangle \subseteq f \langle \tau, \sigma \rangle$ rather simple and I simply wanted to point you to
$$
f \langle \tau, \sigma \rangle \subseteq S_p = \langle f \tau , f \sigma \rangle
$$ is by far even simpler!

I think there might be a problem. $f \langle \tau, \sigma$ is a left-coset and therefore isn't generally a group. So, even though $f \langle \tau, \sigma \rangle$ contains the generators $f \tau$ and $f \sigma$, this doesn't necessarily imply that it contains $\langle f \tau, f \sigma$.

At this point, I am not sure how to fix the proof.

 

[fresh_42]

I think there might be a problem. $f \langle \tau, \sigma$ is a left-coset and therefore isn't generally a group. So, even though $f \langle \tau, \sigma \rangle$ contains the generators $f \tau$ and $f \sigma$, this doesn't necessarily imply that it contains $\langle f \tau, f \sigma$.

At this point, I am not sure how to fix the proof.

I'm not sure what you mean. $f\langle \tau, \sigma \rangle = \{fg\,\vert \,g \in \langle \tau, \sigma \rangle\}$. I haven't checked your "rather simple" part, but as $S_p$ is the entire group, it contains of course its elements $fg$.