There exist one p-ssg then (p-1) elements are of order p. So, if we take q, p-ssg then there are (p-1)q elements are of order p. Similarly, if there are r, q-ssg there are (q-1)r elements are of order q and if pq are r-ssg then there are (r-1)pq elements are of order r. We know 1+(p-1)q + (q-1)r +(r-1)pq ≤ pqr= order => (r-1)(q-1) ≤ 0 But r ≥2, q ≥2 Which is a contradiction. So either q or r SSG is unique. Also we can say G is not simple.
@zero-sl3bn Possible No. of p-ssg 1+pk| qr for k=0,1,2,... are 1,q,r, qr For q SSG 1+qk| pr for k= 0,1,2...... are 1,r, pr For r SSG 1+rk| pq for k=0,1,2.....are 1, pq And here I've mentioned "if we take"
@@zero-sl3bnI have taken 1 case to prove it, we can take another case and check it in the same way. And here assumption is wrong, there is contradiction. And order is p
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Sir, same I was trying with an example of order 30 in the class
There exist one p-ssg then (p-1) elements are of order p.
So, if we take q, p-ssg then there are (p-1)q elements are of order p.
Similarly, if there are r, q-ssg there are (q-1)r elements are of order q and if pq are r-ssg then there are (r-1)pq elements are of order r.
We know 1+(p-1)q + (q-1)r +(r-1)pq ≤ pqr= order
=> (r-1)(q-1) ≤ 0
But r ≥2, q ≥2
Which is a contradiction.
So either q or r SSG is unique.
Also we can say G is not simple.
How do you know there are r q-ssg and q p-ssg ?
@zero-sl3bn Possible No. of p-ssg 1+pk| qr for k=0,1,2,... are 1,q,r, qr
For q SSG 1+qk| pr for k= 0,1,2...... are 1,r, pr
For r SSG 1+rk| pq for k=0,1,2.....are 1, pq
And here I've mentioned "if we take"
@@hrishutiwari4929sorry for previous reply
But it's too tedious
And what about other cases?
Edit: and also what is the order you are taking pr?
@@zero-sl3bn p< q< r
@@zero-sl3bnI have taken 1 case to prove it, we can take another case and check it in the same way. And here assumption is wrong, there is contradiction.
And order is p
HW answer option c