Let be a group with more than one Sylow -subgroup. Over all pairs of distinct Sylow -subgroups let and be chosen so that is maximal. Show that has more than one Sylow -subgroup and that any two distinct Sylow -subgroups of intersect in . (Thus is divisible by and by some prime other than . Note that Sylow -subgroups of need not by Sylow in .)

Note that is proper since is a -group. Since , we have that divides .

Suppose now that is a -group. Then for some Sylow -group . If , then , and hence . However, since and , we have a contradiction since there are elements in and but not in (namely, ). Thus . But then we have and , so that . This implies that is not (cardinality) maximal among the pairwise intersections of Sylow -subgroups. Thus is not a -group, so that its order is divisible by some prime other than .

Suppose now that contains a unique Sylow -subgroup, say . Then . Moreover, for some Sylow -sugbroup of . Since and is order maximal, we in fact have . Similarly, . But then , so that is Sylow in , a contradiction.

Let . Since is a -group, is contained in some Sylow -group . By Sylow’s Theorem, is conjugate to and in , say by and , respectively. Since is normal in , . Thus . Now and for some Sylow -subgroups and of . Suppose . Then, since and are distinct, the subgroup as generated in is a -group, normalizes , and properly contains (for instance) , which is a contradiction since is Sylow in . Thus we have . Moreover, , and since is maximal among the intersections of distinct Sylow -subgroups and is finite, .

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## Comments

And if A’=B’ ?

Good question.

I’ll mark this “incomplete” until I have a chance to fix it. (Or until someone offers a fix.)

Thanks!

OK- I think it’s patched up now.

Let me know if you find another problem, and thanks for the heads up!

No, it looks ok.

Can you please explain more clearly “why N_G(P n Q) is contained in P” in the 4th line ?thnx

That’s actually not necessarily true; fortunately it’s also not what I meant.🙂 It now reads and , so that .

Thanks for letting me know!

thnx.(P n Q) is strictly contained in (P n R) but i think this is not a contradiction.it just implies that P=R.then we may possibly derive a contradiction but i couldnt find out what?

The contradiction is that was chosen to have maximal cardinality among the pairwise intersections of Sylow -subgroups in . If , then there is a larger intersection.

Oops- I see what you mean now.

Hmm…

Okay, I think it’s fixed.

If , then we get that is contained in both and but strictly contains , which is a contradiction.

Thanks for reading so closely! Let me know if you find another problem.