0
$\begingroup$

I was reading this, question (shown in pic). But I didn't get, in (3) why they doesn't consider case of $|Q_i ∩ Q_j| = 9$ ?

Since, $Q_i ∩ Q_j ≤ Q_i$

$→ |Q_i ∩ Q_j| = 9$ is also valid case and if $|Q_i ∩ Q_j| = 9$ then what happens?

first page of proof

second page of proof.

$\endgroup$
1
$\begingroup$

If $|Q_i\cap Q_j|=9$, then $Q_i=Q_j$ (and so $i=j$), since $|Q_i|=|Q_j|=9$. Since the cases (a) and (b) in the argument are based on the intersections $Q_i\cap Q_j$ for $i\neq j$, this possibility is thus irrelevant.

$\endgroup$
  • $\begingroup$ How $Q_i = Q_j$ ? Since there are two groups of order 9 upto isomorphism. Order are equal doesn't mean groups are equal? $\endgroup$ – Akash Patalwanshi Nov 25 '17 at 6:48
  • $\begingroup$ Well, $Q_i=Q_i\cap Q_j=Q_j$, since $Q_i\cap Q_j$ contains all $9$ elements of both of them... $\endgroup$ – Eric Wofsey Nov 25 '17 at 6:50
  • $\begingroup$ Yes. Thanks sir. but if "$Q_i = Q_j$ then what? How the possibility of 9 discarded? $\endgroup$ – Akash Patalwanshi Nov 25 '17 at 6:52
  • $\begingroup$ Thanks sir got it.. $\endgroup$ – Akash Patalwanshi Nov 25 '17 at 6:58

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.