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?
Best Answer
$|G|=3\cdot5\cdot7$
$$ n_5 \in\{1, 21\} \; \text{&} \; n_7 \in \{1,15\}$$
$$ \text{If} \; n_5>1 \; \text{and} \; n_7>1 $$
$$\Rightarrow n_5=21 \; \text{&} \; n_7=15 $$
$$\Rightarrow \text{There are } \; 21(5-1) + 15(7-1) >105\; \text{elements of order 5 and 7 together, which is absurd.} $$
$$ \text{Hence either } \; n_5=1 \; \text{or} \; n_7=1$$
$$\text{If P and Q are Sylow 5 and 7 subgroups respectively then, one of the two has to be normal in G}$$
$$\text{Without loss of generality, assume P is normal in G }$$
$$P\lhd G \;\text{and} \;Q \leq G \Rightarrow PQ \leq G $$
$$\;\text{Futhermore}, P\cap Q=\{e\} \Rightarrow |PQ|=\frac{|P||Q|}{|P\cap Q|}=35\; $$
$$[G:PQ]=3, \;\text{which also happens to be the least prime dividing order of G, hence}\; PQ\lhd G$$
$$|PQ|=5.7 \; \text{ and 5 doesn't divide (7-1), so } \; PQ\; \text{is cyclic} $$
$$\Rightarrow \;\text{P and Q are characteristic in PQ. Since PQ is normal in G, we have both P and Q normal in G}$$