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.


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.

  • $\begingroup$ 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

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