Question

Suppose that \(f\) is some polynomial of prime order \(p\) over the rationals and that it has exactly two nonreal roots. I just need one idea to show that the Galois group of its splitting field is isomorphic to the symmetric group of \(p\) letters.

Answer 1

I've so far been able to show that \(x\mapsto\overline x\) is in \(\text{Gal}\left(K/\mathbb Q\right)\), where \(K\) is the splitting field of \(f\), and has order \(2\). Also, if \(\alpha\) is one of those nonreal roots, then \(\left[\mathbb Q\left(\alpha\right):\mathbb Q\right]=p\), which implies that \(\text{Gal}\left(K/\mathbb Q\right)\) has an element of order \(p\) by Cauchy's theorem.

Answer 2

I may have read that if \(\alpha\) is a \(2\)-cycle and \(\beta\) is a \(p\)-cycle (prime), then \(\langle\alpha,\beta\rangle\) generate the symmetric group of \(p\) letters. Is this even true?

Answer 3

I think I just need to show that a \(2\)-cycle and a \(p\)-cycle, where \(p\) is prime, generate \(S_p\), if anyone might want to help me with that.

Answer 4

@LetsLearn2000, my name is Susanne. Pleasure to meet you.

Answer 5

Im Vanessa. Nice to meet you as well Susanne.

Answer 6

I've been a member of this site for five years. It's all new faces now. T_T All of my friends are gone!

Answer 7

yeah, its a lot of people just like you who have been on here for years, and some people are new.

Answer 8

I don't recognize half of the guys here.

Answer 9

yeah, its surely some new faces lol. Could you continue helping out with those questions?

Answer 10

@ganeshie8 @Kainui Hardcore stuff that is usually not seen on here.

Answer 11

Nope, haven't taught myself finite fields yet, it's on my todo list though.

Answer 12

It's out of my knowledge as well... but I do know that every group of prime order is cyclic... if that helps at all :P

Answer 13

What I can gather from the internet is that a 2-cycle that swaps adjacent elements and a n-cycle together generates the symmetric group of order n.

Answer 14

*symmetric group of n letters, not order n. I am not familiar with group theory at all.

Answer 15

@thomas5267, so the \(n\)-cycle need not be prime. Hmm. I'm still struggling to prove this.

Answer 16

If I prove this, I will be able to say that \(S_p\cong\langle\tau,\sigma\rangle=\text{Gal}\left(K/\mathbb Q\right)\) since \(\left[K:\mathbb Q\right]\) can be at most \(p!\), the order of \(S_p\).

Answer 17

What is that 2-cycle in question? Did you prove a 2-cycle exists or were you able to explicitly construct such 2-cycle?

Answer 18

I suspect since the symmetric group has p letters, it might be the case that any 2-cycle is good enough?

Answer 19

It's the first sentence of my first post.

Answer 20

Never mind. I got it. I reduced the problem to showing that the set of transpositions of \(S_n\) of the form \(\left(i,i+1\right)\), where \(0\leqslant i< n-1\), generates \(S_n\). Then I will use the fact that
\[\left(1,2,\dots,n\right)^i\left(1,2\right)\left(1,2,\dots,n\right)^{-i}=\left(i+1,i+2\right).\]