Monthly Archives: May 2010

Every group of order 315 with a normal Sylow 3-subgroup is abelian

Let G be a group of order 315 which has a normal Sylow 3-subgroup P_3. Prove that P_3 \leq Z(G) and that G is abelian.


Note that 315 = 3^2 \cdot 5 \cdot 7. The divisibility and congruence criteria of Sylow’s Theorem force n_5 \in \{ 1,21 \} and n_7 \in \{ 1,15 \}.

Note that G/C_G(P_3) \leq \mathsf{Aut}(P_3), where \mathsf{Aut}(P_3) is a group of order 6 or 48 by the comments at the end of §4.4. Hence G/C_G(P_3) = 1, so that P_3 \leq Z(G).

Consider now G/P_3; this is a group of order 35. By Sylow’s Theorem, G/P_3 has a unique Sylow 5-subgroup Q_5. Moreover, each element of Q_5 has 3 preimages under the natural projection \pi : G \rightarrow G/P_3, for a total of 3 \cdot 5 = 15 elements in G that map to Q_5.

Let P_5 be a Sylow 5-subgroup of G. Now P_5 \not\leq P_3, so that \pi[P_5] \neq 1. Note then that \pi[P_5] \leq G/P_3 is a subgroup of order 5; hence \pi[P_5] = Q_5. If n_5 = 21, then at least 4 \cdot 21 = 84 elements of G are mapped to Q_5 by \pi, a contradiction. Thus n_5 = 1, so that P_5 is unique (as a Sylow 5-subgroup) and normal.

Now G/C_G(P_5) \leq \mathsf{Aut}(P_5). Since |\mathsf{Aut}(P_5)| = 4, G/C_G(P_5) = 1, so that C_G(P_5) = G and we have P_5 \leq Z(G).

Now \langle P_3,P_5 \rangle \leq Z(G) \leq G. By Lagrange’s Theorem, |Z(G)| is either 45 or 315.

If |Z(G)| = 45, then G/Z(G) is cyclic, so that G is abelian. Hence Z(G) = G, a contradiction. Thus |Z(G)| = 315, and we have G = Z(G). Thus G is abelian.

If a group of order 105 has a normal Sylow 3-subgroup, then the group is abelian.

Let G be a group of order 105. Prove that if a Sylow 3-subgroup of G is normal then G is abelian.


Note that 105 = 3 \cdot 5 \cdot 7. We showed in §4.5 #17 that n_5 = n_7 = 1; suppose also that n_3 = 1. Let P_3, P_5, and P_7 denote the (unique hence normal) Sylow 3-, 5-, and 7-subgroup of G, respectively.

By Corollary 15 in §4.4, G/C_G(P_3) \leq \mathsf{Aut}(P_3). Since |\mathsf{Aut}(P_3)| = 2, we have G/C_G(P_3) = 1, so that P_3 \leq Z(G).

Similarly, G/C_G(P_5) \leq \mathsf{Aut}(P_5), which is a group of order 4. Thus G/C_G(P_5) = 1, so that P_5 \leq Z(G).

Now \langle P_3, P_5 \rangle \leq Z(G) \leq G. By Lagrange, |Z(G)| is either 15 or 105.

If |Z(G)| = 15, then G/Z(G) \cong Z_7 is cyclic. Hence G is abelian, and we have |Z(G)| = 105, a contradiction.

Thus |Z(G)| = 105, and we have G = Z(G). Thus G is abelian.

Every group of order 385 contains a central Sylow 7-subgroup and a normal Sylow 11-subgroup

Prove that if G is a group of order 385, then Z(G) contains a Sylow 7-subgroup of G and a Sylow 11-subgroup is normal.


Note that 385 = 5 \cdot 7 \cdot 11. Now Sylow’s Theorem forces n_7 = n_{11} = 1, so that G has unique (hence normal) Sylow 7- and Sylow 11-subgroups. Let P_7 denote the (unique) Sylow 7-subgroup.

We have N_G(P_7) = G. By the N/C Theorem, G/C_G(P_7) \leq \mathsf{Aut}(P_7). Moreover, |\mathsf{Aut}(P_7)| = 6 = 2 \cdot 3. Thus |G/C_G(P_7)| = 1, so that G/C_G(P_7) = 1, hence C_G(P_7) = G. Thus P_7 \leq Z(G).

Every group of order 231 has a central Sylow 11-subgroup and a normal Sylow 7-subgroup

Prove that if G is a group of order 231 then Z(G) contains a Sylow 11-subgroup of G and a Sylow 7-subgroup is normal.


Note that 231 = 3 \cdot 7 \cdot 11. The divisibility and congruence criteria of Sylow’s Theorem force n_7 = n_{11} = 1, so that G has unique (hence normal) Sylow 7- and Sylow 11-subgroups; let P_{11} be the unique Sylow 11-subgroup.

By the N/C Theorem, G/C_G(P_{11}) \leq \mathsf{Aut}(P_{11}). By Proposition 16 in §4.4, |\mathsf{Aut}(P_{11})| = 10 = 2 \cdot 5; thus |G/C_G(P_{11})| = 1, so that G/C_G(P_{11}) = 1, hence C_G(P_{11}) = G. Thus P_{11} \leq Z(G).

No simple group of order 462 exists

Prove that if |G| = 462 then G is not simple.


Note that 462 = 2 \cdot 3 \cdot 7 \cdot 11. By Sylow’s Theorem, n_{11} is congruent to 1 mod 11 and divides 42. The only integer satisfying these conditions is 1; thus G has a unique (hence normal) Sylow 11-subgroup. So G is not simple.

No simple group of order 132 exists

Prove that if |G| = 132 then G is not simple.


Note that 132 = 2^2 \cdot 3 \cdot 11. Suppose G is simple.

The divisibility and congruence criteria of Sylow’s Theorem force the following.

  • n_2 \in \{ 1,3,11,33 \}
  • n_3 \in \{ 1,4,22 \}
  • n_{11} \in \{ 1,12 \}

Since G is simple, we have n_{11} = 12, so that G has 12 \cdot 10 = 120 elements of order 11. If n_3 = 22, then G has at least 120 + 22 \cdot 2 = 164 elements, a contradiction. So n_3 = 4. There are only 4 remaining elements, which must comprise a Sylow 2-subgroup, which is unique and thus normal- a contradiction.

Thus no group of order 132 is simple.

No simple group of order 2907 exists

Prove that if |G| = 2907, then G is not simple.


Note that 2907 = 3^2 \cdot 17 \cdot 19. The divisibility and congruence criteria of Sylow’s Theorem force the following.

  • n_3 \in \{ 1,19 \}
  • n_{17} \in \{ 1,171 \}
  • n_{19} \in \{ 1,153 \}

Suppose G is simple. Then n_3 = 19, n_{17} = 171, and n_{19} = 153. The Sylow 17- and 19-subgroups of G are cyclic and thus intersect trivially. Thus G has at least 153 \cdot 18 + 171 \cdot 16 = 5490 elements, a contradiction.

Thus G is not simple.

No simple group of order 1365 exists

Prove that if |G| = 1365 then G is not simple.


Note that 1365 = 3 \cdot 5 \cdot 7 \cdot 13. The divisibility and congruence criteria of Sylow’s Theorem force the following.

  • n_3 \in \{ 1,7,13,91 \}
  • n_5 \in \{ 1,21,91 \}
  • n_7 \in \{ 1,15 \}
  • n_{13} \in \{ 1,105 \}

Suppose G is simple. Then n_p \neq 1 for p \in \{ 3,5,7,13 \}. Because the Sylow subgroups of G have prime order, they are cyclic, and so intersect trivially. Thus G has at least 105 \cdot 12 + 15 \cdot 6 + 21 \cdot 4 + 7 \cdot 2 = 1448 elements, a contradiction.

Thus G is not simple.

No simple group of order 6545 exists

Let G be a group. Prove that if |G| = 6545 then G is not simple.


Note that 6545 = 5 \cdot 7 \cdot 11 \cdot 17. Note that the divisibility and congruence criteria of Sylow’s Theorem force the following.

  • n_5 \in \{ 1, 11 \}
  • n_7 \in \{ 1, 85 \}
  • n_{11} \in \{ 1, 595 \}
  • n_{17} \in \{ 1, 35 \}

Note in particular that for each prime p dividing 6545, the Sylow p-subgroups of G intersect trivially. If no Sylow subgroup is normal, then G has at least 4 \cdot 11 + 6 \cdot 85 + 10 \cdot 595 + 16 \cdot 35 = 7094 elements, a contradiction. Thus some Sylow subgroup of G is normal, and hence G is not simple.

Every group of order 200 has a normal Sylow 5-subgroup

Prove that a group of order 200 has a normal Sylow 5-subgroup.


Note that 200 = 2^3 \cdot 5^2. By Sylow’s Theorem, n_5 is congruent to 1 mod 5 and divides 8; thus n_5 = 1, and so the (unique) Sylow 5-subgroup of G is normal.