2
$\begingroup$

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, $$G \cong \mathbb{Z}_{pqr}$$

$\endgroup$
  • 1
    $\begingroup$ The result is definitely true and can be easily proven $\endgroup$ – Learnmore Oct 14 '17 at 2:42
1
$\begingroup$

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 $HK\times L\cong \Bbb Z_{pq}\times \Bbb Z_r\cong \Bbb Z_{pqr}$

$\endgroup$
  • $\begingroup$ This exactly my argument. $\endgroup$ – Ahmed Elashry Oct 14 '17 at 5:07
0
$\begingroup$

A finite group is nilpotent iff. every Sylow subgroup is normal. Thus $G$ is nilpotent, of squarefree order. It follows immediately that $G$ is cyclic since all of its Sylow subgroups are cyclic.

What you've tried to do is to show the internal product of the Sylow subgroups is isomorphic to the external direct product. That follows immediately from the trivial intetsection of pairwise Sylow subgroups and their normality. Use induction to generalize this to groups of any squarefree order.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.