# Group of Order 105 Having Normal Sylow 5/7 Subgroups

I am trying to prove that, given $|G|=105$, the G has a normal Sylow 5 subgroup and a normal Sylow 7 subgroup.

I think the thing that is confusing me is the word "and". It would seem that there can't be both a normal subgroup of order 5 and 7, i think. If there were, since their intersection is just $e_G$, then this ties up the remaining 94 elements in a bunch of Sylow 3 subgroups. Again, because they each contain $e_G$, there needs to be 47 Sylow 3 subgroups, which is impossible by the conditions on the Sylow theorems; So where is my thinking off? Does it really mean "or"?

There has to be at least one normal 5 or 7, since if neither were normal, because $n_5=21$ and $n_7=15$, and since $4(21)+6(15)>105$, we have a contradiction with the number of elements. So where is my logic flawed?

Fact 1: If $|G|=pqr$ with $p>q>r$, then Sylow-$p$ subgroup is normal.

Fact 2: (Schur-Zassenhaus) If $H\leq G$ such that $|H|$ and $|G\colon H|$ are coprime, then $G$ contains subgroup of order $|G\colon H|$.

If $|G|=105$, Sylow-$7$ subgroup, say $P$, must be normal. Then $|P|$ and $|G\colon P|$ are co-prime, hence $G$ contains subgroup of order $|G\colon P|=15$, say $Q$, and $G=P\rtimes Q$.

Then $Q$ is cyclic group of order $15$, and acts by conjugation on $P$, giving a homomorphism from $\mathbb{Z}_{15}\cong Q\rightarrow Aut(P)\cong \mathbb{Z}_6$. In this homomorphism, the elements of order $5$ in $Q$ must go to identity in $Aut(P)$ since there are no elements of order $5$ in $Aut(P)$.

This implies, the subgroup of order $5$ in $Q$ (Sylow-$5$ subgroup), acts trivially by conjugation on Sylow-$7$ subgroup $P$; i.e. it commutes.

Thus, Sylow-$5$ subgroup commutes with Sylow-$3$ subgroup as well as Sylow-$7$ subgroup. This means Sylow-$5$ subgroup is in center of $G$, and hence $$G=(\mbox{Sylow-5 subgroup})\times (\mbox{subgroup of order 21}).$$

From this, we can see that both Sylow-$5$ and Sylow-$7$ are normal.

I have used two non-trivial facts from group theory. But, once we ensure that both Sylow-$5$ and $7$ are normal, one may try to give different, elementary argument for proof. Since you were worrying about and/or (both appearing in question title and first line of question), this was just to ensure that it is and.)

$$|G|=3\cdot5\cdot7$$

$$n_5 \in\{1, 21\} \; \text{&} \; n_7 \in \{1,15\}$$

$$\text{If} \; n_5>1 \; \text{and} \; n_7>1$$

$$\Rightarrow n_5=21 \; \text{&} \; n_7=15$$

$$\Rightarrow \text{There are } \; 21(5-1) + 15(7-1) >105\; \text{elements of order 5 and 7 together, which is absurd.}$$

$$\text{Hence either } \; n_5=1 \; \text{or} \; n_7=1$$

$$\text{If P and Q are Sylow 5 and 7 subgroups respectively then, one of the two has to be normal in G}$$

$$\text{Without loss of generality, assume P is normal in G }$$

$$P\lhd G \;\text{and} \;Q \leq G \Rightarrow PQ \leq G$$

$$\;\text{Futhermore}, P\cap Q=\{e\} \Rightarrow |PQ|=\frac{|P||Q|}{|P\cap Q|}=35\;$$

$$[G:PQ]=3, \;\text{which also happens to be the least prime dividing order of G, hence}\; PQ\lhd G$$

$$|PQ|=5.7 \; \text{ and 5 doesn't divide (7-1), so } \; PQ\; \text{is cyclic}$$

$$\Rightarrow \;\text{P and Q are characteristic in PQ. Since PQ is normal in G, we have both P and Q normal in G}$$