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?