0
$\begingroup$

My question is: why does $$A_\infty=\bigcup_{n\geq 5}^{\infty} \,A_n \,\,?$$

Doesn't $A_{\infty}$ contain (an isomorphic copy of) $A_4$, which is not a simple group? I'm sorry if this is a stupid question, I just haven't been able to see it. This comes from here: Is the question phrased properly? and is my proof correct? (An infinite alternating group is simple).

$\endgroup$
  • 2
    $\begingroup$ Well, $A_{5}$ already contains $A_{4}$. $\endgroup$ – Geoff Robinson Oct 20 '15 at 21:47
  • $\begingroup$ I think I see. Does that mean that $A_\infty = \bigcup_{n\geq k}^{\infty} A_n$ for any $k\geq 5$? I feel my confusion came from not understanding why 5 was chosen in particular. $\endgroup$ – user282311 Oct 20 '15 at 21:51
  • 1
    $\begingroup$ Yes because you can show that $A_{n-1}<A_n$. $A_5$ is the smallest simple alternating group (and also smallest simple non-abelian group). $\endgroup$ – Ali Caglayan Oct 20 '15 at 21:53
  • $\begingroup$ Starting from 5 is to emphasize that $A_\infty$ is a union of finite simple groups. $\endgroup$ – Amin Nov 18 '15 at 9:54

Your Answer

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

Browse other questions tagged or ask your own question.