I was trying to solve a problem about how many groups of order $1001$ using Sylow's theorems. I proved that there is only one, and noticed that the argument can be generalized to a statement like:
Let $G$ be a group of order pqr, where p,q, and r are different primes. If every Sylow subgroup is normal, then G is abelian.
I searched the internet, but I did not find a proof or a counterexample, so I need to check my argument.
Proof:
Let $H_p$, $H_q$, and $H_r$ be the Sylow subgroups of $G$.
Consider $J = H_p H_q H_r \subset G$.
Since all the Sylow subgroups are normal in $G$, then they are normal in $J$.
The Sylow subgroups are of prime order, then they intersect trivially and hence they commute. For, if $x$ and $y$ in different Sylow subgroups, we have
$$ xyx^{-1}y^{-1} = (xyx^{-1})y^{-1} = x(yx^{-1}y^{-1})$$
Hence, $xyx^{-1}y^{-1}$ in the intersection, and $xyx^{-1}y^{-1} = e$, then $xy = yx$.
We have $$K = H_p H_q \triangleleft J $$
To prove this we just need to look on elements of $H_r$.
$\forall g \in H_r$, we have
$$ gH_p H_q g^{-1} = H_p gg^{-1}H_q = H_p H_q$$
This gives $K \cong H_p \times H_q \cong \mathbb{Z}_p \times \mathbb{Z}_q \cong \mathbb{Z}_{pq} $.
Now, $J = KH_r$ , $K \triangleleft J$ , and $H_r \triangleleft J$, then
$$ J \cong K \times H_r \cong \mathbb{Z}_{pq} \times \mathbb{Z}_r \cong \mathbb{Z}_{pqr} $$
Hence, $|J| = |\mathbb{Z}_{pqr}| = 1001$, and then $J = G$. Thus,
$$G \cong \mathbb{Z}_{pqr}$$
Best Answer
Suppose the subgroups are $H,K,L$ of order $p,q,r$ respectively .Then since each $H,K,L$ are unique Sylow Subgroups of $G$ they are normal in $G$.
Now consider $HK$ .Since $H,K$ are normal in $G$ hence $HK$ is a normal subgroup of $G$ of order $pq$. Now $H$ is a cyclic group of order $p$ and $K$ is a cyclic group of order $q$.
Since $H\cap K=\{e\}\implies HK\cong H\times K\cong \Bbb Z_p\times \Bbb Z_q \cong \Bbb Z_{pq}$
Now consider $HK,L$ and each of them form a normal subgroup of $G$.
Consider the subgroup $HKL$ which will be isomorphic to $HK\times L\cong \Bbb Z_{pq}\times \Bbb Z_r\cong \Bbb Z_{pqr}$