## 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.