Show that no group of order p^2 q is simple, where p and q are distinct primes.
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☆☆heya your answer is here☆☆
SOLUTION;-
First, suppose p<qp<q. Then nq≡1(modq)nq≡1(modq), and nq∣p2nq∣p2. So the only possibilities are that nq=1,p,p2nq=1,p,p2. If nq=1nq=1, the group is not simple. But nq=pnq=p is impossible, else q∣p−1q∣p−1. So nq=p2nq=p2. Then we have p2p2 groups of prime order qq, so they have pairwise trivial intersection, and this forces the pp-Sylow subgroup to be unique.
The qq-Sylow subgroups give a total of p2(q−1)p2(q−1) elements of order qq. This accounts for all but p2p2 elements of the group, so any pp-Sylow subgroup must be these remaining elements, hence unique.
The case where q<pq<p is quite quick. Any pp-Sylow subgroup has index qq. This is smallest prime dividing |G||G|, so your pp-sylow subgroup is normal.
SOLUTION;-
First, suppose p<qp<q. Then nq≡1(modq)nq≡1(modq), and nq∣p2nq∣p2. So the only possibilities are that nq=1,p,p2nq=1,p,p2. If nq=1nq=1, the group is not simple. But nq=pnq=p is impossible, else q∣p−1q∣p−1. So nq=p2nq=p2. Then we have p2p2 groups of prime order qq, so they have pairwise trivial intersection, and this forces the pp-Sylow subgroup to be unique.
The qq-Sylow subgroups give a total of p2(q−1)p2(q−1) elements of order qq. This accounts for all but p2p2 elements of the group, so any pp-Sylow subgroup must be these remaining elements, hence unique.
The case where q<pq<p is quite quick. Any pp-Sylow subgroup has index qq. This is smallest prime dividing |G||G|, so your pp-sylow subgroup is normal.
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hy.
here is your answer
First suppose that q < p. Then a p-Sylow subgroup of G has
index the smallest prime dividing |G|, and hence is normal by problem 12 of
section 7.3. So suppose that q > p. Recall that the number nq of q-Sylow
subgroups Q is congruent to 1 modulo q and divides the index of Q in |G|,
which in this case is ²2
. So the only possibilities are 1, p, and p²
. If nq = 1,
Q is normal, and we are done. Since p < q, we can’t have p ≡ 1 (mod q).
If nq = p²
, there are p²
subgroups of order q, and their only intersection is in
the identity. This gives us p²
(q − 1) elements of exact order q, leaving only
p²
elements remaining in the group. But any p-Sylow subgroup P must then
consist of all these remaining elements. This implies that P is unique, hence
normal, so the again G is not simple.
here is your answer
First suppose that q < p. Then a p-Sylow subgroup of G has
index the smallest prime dividing |G|, and hence is normal by problem 12 of
section 7.3. So suppose that q > p. Recall that the number nq of q-Sylow
subgroups Q is congruent to 1 modulo q and divides the index of Q in |G|,
which in this case is ²2
. So the only possibilities are 1, p, and p²
. If nq = 1,
Q is normal, and we are done. Since p < q, we can’t have p ≡ 1 (mod q).
If nq = p²
, there are p²
subgroups of order q, and their only intersection is in
the identity. This gives us p²
(q − 1) elements of exact order q, leaving only
p²
elements remaining in the group. But any p-Sylow subgroup P must then
consist of all these remaining elements. This implies that P is unique, hence
normal, so the again G is not simple.
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