Comments on: A Group of Order $pqr$ Contains a Normal Subgroup of Order Either $p, q$, or $r$

By: Lowly Undergraduate

Lowly Undergraduate — Mon, 11 Nov 2019 13:34:09 +0000

I made the same observation. I believe he means the line

(r−1)pq + (q−1)r + (p−1)q = pqr + qr − r − q

should instead be

(r−1)pq + (q−1)r + (p−1)q + 1 = pqr + qr − r − q + 1.

Indeed, I agree you may omit the identity element when counting the elements of the Sylow p-subgroups, but in the end, I think you need to add it back in. This then gives us q=3, so p=2 and r=5. The result is straightforward from here.

Please correct me if I misunderstand.

Sincerely,

a lowly undergraduate

By: Yu

Yu — Fri, 04 Oct 2019 23:02:09 +0000

Dear Xiaoqi Wei,

I think I don’t have to count the identity element because we need to count different elements. Please let me know if I didn’t get your point.

By: Xiaoqi Wei

Xiaoqi Wei — Mon, 30 Sep 2019 18:25:54 +0000

I think you may forget to count identity element.

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