1
$\begingroup$

How to show $\langle x, y\rangle=S_7$? For arbitrary $2$ cycle $x$ and $7$ cycle $y$ in $S_7$?

So far I have by Lagrange's that $2 | \langle x,y\rangle$ and $7|\langle x,y\rangle$ and $\langle x,y\rangle|S_7$.

In other questions of a similar nature I have had success by doing case by case and ruling out all possibilities but here since $|S_7|=7!=5040$ I don't see that as being very helpful.

Any ideas?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ This is related: math.stackexchange.com/a/357673/160300 $\endgroup$ – Crostul May 10 '17 at 11:15
  • $\begingroup$ Can you tell us what your angle-brackets mean? If the question, it appears that they represent some group or subgroup (else they could not equal $S_7$). In the text, you have $2$ dividing an angle-bracket thing, suggesting it's an integer. Clarify, please? $\endgroup$ – John Hughes May 10 '17 at 11:15
  • $\begingroup$ I guess it's the generating set of the group $S_7$ $\endgroup$ – Arpan1729 May 10 '17 at 11:34
1
$\begingroup$

$(1,2)$ and $(1,2,...,n)$ generate $S_n$. Conclude from this that $(i,j)$ and $(1,2,...,n)$ generate $S_n$ whenever $gcd(i-j,n)=1$. Your problem is solved.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.