# Proof that every group of order 847 is abelian.

How would I use Sylow's theorems to prove that any group of order 847 is abelian? I know that every group of prime order is cyclic, and thus, abelian. However, $847 = 7*11*11$, thus, is not a prime number.

• If I'm not mistaken, both Sylow subgroups are normal. That makes this an easy one. – Matt Samuel Dec 2 '16 at 21:36
• Are Sylow subgroups normal in general? – PiccolMan Dec 2 '16 at 21:56
• If they were always normal the Sylow theorems would be a lot less useful. – Matt Samuel Dec 2 '16 at 21:57

## 1 Answer

Let $n_p$ denote the number of $p$-Sylow subgroups of $G$. We are interested in $n_7$ and $n_{11}$. We know that

1. $n_7\mid 121$, $n_7\equiv 1\pmod{7}$

2. $n_{11}\mid 7$, $n_{11}\equiv 1\pmod{11}$

It follows that $n_7=n_{11}=1$, can you see why? If $A$ is the $7$-Sylow subgroup and $B$ is the $11$-Sylow subgroup, we have $G\cong A\times B$.

Can you finish?

• Oh, I see. Every finite abelian group is the product of cyclic groups. Also, a group is cyclic if it is of prime order. So we must show that $G\cong A\times B$, where $A$ and $B$ are subgroups of $G$ by using the internal direct product. In order to use the internal direct product $A \cap B = \{ e \}$, $A$ and $B$ must be abelian with respect to each other, and $AB = G$. $A$ can be a Sylow 7-subgroup and $B$ can be a Sylow 11-subgroup. Since each element in $A$ is some power of $7$ and each element of $B$ is some power of $11$ then $A \cap B = {e}$. How would I show that $AB = G$? – PiccolMan Dec 2 '16 at 21:54
• Also, how would I show that $A$ and $B$ are abelian with respect to each other? – PiccolMan Dec 2 '16 at 21:55
• @PiccolMan It's a general result that if all Sylow subgroups are normal then the group is the direct product of them. Now $A$ is clearly abelian; what about $B$? – egreg Dec 2 '16 at 21:56
• I know that every group of order $p^{2}$ is abelian, but I don't know why. $|B| = 11^{2}$, thus, $B$ is abelian. But why is this true? – PiccolMan Dec 2 '16 at 22:55
• @PiccolMan Well, $B\cong G/A$. – egreg Dec 2 '16 at 22:58