Let G be a group with |G|=p²q, where p&q are distinct primes, show that G has a normal Sylow p-subgroup or?

a normal Sylow q-subgroups.

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  • Anonymous
    1 decade ago
    Favorite Answer

    If q < p, then the number of Sylow p-groups is = 1 (mod p), and divides q. Of course then, the number is 1, which means the Sylow p-group is normal in G.

    If p < q, then the number of Sylow q-groups = 1 (mod q) and divides p^2. Therefore, since p<q, either there is one Sylow q-group, or p^2 many. If there is one, we are done. If there are p^2, then there are p^2 * (q-1) elements of order q in G, which means there are |G| - p^2 * (q-1) = p^2 elements in G that are not of order q. These elements must lie in a Sylow p-group, and since there are exactly p^2, there can only be one Sylow p-group, and thus it is normal.

    Steve

  • 4 years ago

    because |G:N_P(G)| = 168 = 2^3 * 3 * 7, the Sylow Theorems propose that there are a million or 8 such 7-Sylow subgroups of G:N_P(G). If there is just one such subgroup, then G:N_P(G) is standard in G:N_P(G). If there are 8 such subgroups, then [f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778] = 8. What we would desire to coach is that if we glance at <f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778f127e572bfb4a45ea5b9c529a2c55778, t>, the place t is a few nontrivial ingredient of G:N_P(G) no longer of order 7, then we generate all of G:N_P(G). extra approximately this quickly... Editf127e572bfb4a45ea5b9c529a2c55778 f127e572bfb4a45ea5b9c529a2c55778o new strategies in this. If I do think of of something, i'm going to aid you comprehend later. in the propose time, reliable success!

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