Youre absolutely right I screwed up. I realised

this about an hour after my post. So of course the

proof is invalid. Here is how i think it will go

:<br><br>Using a bit more theory of Sylow p-subgroups, one shows

that IF there are 7 Sylow 2-subgroups then there is

only one Sylow 7-subgroup (there are either 1 or 8 of

these). I dont think i can show that the case of one

Sylow 2-subgroup is impossible. Similarly, i think that

on the assumption that there are 8 Sylow 7-subgroups

then there must be only one Sylow 2-subgroup. Geez I

can blab on. What I mean to say is that some groups

of order 56 have only one Sylow 2-subgroup and some

have only one Sylow 7-subgroup but they ALL have

either only one Sylow 2-subgroup or only one Sylow

7-subgroup. I havent proven this, Im just saying I think it

is how the proof should go.