# No group of order 2907 is simple

Show that any Group of order 2907 is not a simple group? 2907= 3*3*17*19 I've started with the Sylow 19-subgroup, then the 17-subgroups and finally the 3-subgroups but i couldn't proceed in the proof to find the nontrivial normal subgroup! Please help.

• Can you show your work? – Tim Ratigan Nov 27 '13 at 20:11
• @Tim.Ratigan i'm not much familiar with writing using mathematics symbols ! What i did is that I found the number of each sylow subgroup, but couldn't find a way to find a normal one. – Enas Nov 27 '13 at 20:24
• This tutorial might help. – Tim Ratigan Nov 27 '13 at 20:37
• @Tim.Ratigan thank u :) but i prefer to finish my studying for my midterm exam now, then i'll be ready to learn anything else! – Enas Nov 27 '13 at 20:46

Suppose the $19$-Sylow subgroup(s) isn't(aren't) normal. Are there enough elements left over for the $17$-Sylow subgroup(s) to not be normal? And vice versa.
• If the $17$-Sylow subgroups aren't normal, how many of them are there? – Daniel Fischer Nov 27 '13 at 20:20
• Right. If the $19$-Sylow subgroups aren't normal, there isn't enough space left for $171$ $17$-Sylow subgroups, so the $17$-Sylow subgroup then must be normal. – Daniel Fischer Nov 27 '13 at 20:34