verification of my proof that $|G|=35$ has normal subgroups of orders 5 and 7.

Question

let $|G|=35$.

I want to show that G has a normal subgroup of order 5 and a normal subgroup of order 7.

My attempt:

$|G|=35=5^17=7^15.$

$n_5( \text{the number of 5-sylow groups})=(1+5k)|7$. But only 7 and 1 divide 7 and $1+5k\neq 7 \forall k \geq 0$. So we conclude that $n_5=1$

Now we can note that $|gHg^{-1}|=|H|$ so Therefore considering that we know the 5 sylow group is a subgroup of G we know that $|gPg^{-1}|=|P|$ and seems as how this is the only subgroup of this order it implies that $gPg^{-1}=P, \forall g\in G$ and so P is normal.

Similarly for $n_7$

$n_7=(1+7k)|5$ which implies (for the same reasons as above ) that k=0 and $n_7=1$

We now use an identical argument to the one above to show that P is normal here aswell

Note: I have used P for both the case of the 5 sylow subgroup and the 7-sylow subgroup but the context is obvious I believe.

Answer 1

The proof is correct. Here are some notes:

• the transition: Sylow $p$-subgroup of a finite group $G$ is unique $\Rightarrow$ it is normal in $G$ is good to remember, because it is easy and yet very critical in most cases.
• as a consequence of the proof we have got that any group of order $35$ is isomorphic to $C_{35}$ - cyclic group. There is a (not easy and unobvious) generalization: any group of order $n$ is cyclic iff $n$ and $\varphi(n)$ are coprime.