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Show that a group of order $70$ can not be simple.

I've tried to solve using Sylow theorem. I got $1, 5, 7, 35$ Sylow $2$-subgroups, $1$ sylow $5$-subgroup and $1$ sylow $7$-subgroup. Now the only choice is $35$ Sylow $2$-subgroups which would yield $36$ elements. Now we are left with $34$ elements but we have only one sylow $5$-subgroup and one sylow $7$-subgroup.

Why all the elements of sylow subgroups are not adding up to $70$?

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Probably because your group might have elements of order 14 for example, which don't belong to any Sylow subgroup... – N. S. Apr 6 '12 at 19:03
Probably there are elements of composite order. – user21436 Apr 6 '12 at 19:04
BTW, you may want to edit your title. Your question seems to be about something connected with but not exactly about simpleness of groups. – user21436 Apr 6 '12 at 19:06
you know that 70=2.5.7 so use this fact – Babak Miraftab Apr 6 '12 at 19:07
up vote 5 down vote accepted

The elements in a Sylow subgroup have a prime power as order, if the group has an element of order 35 (for example) it wont be in a sylow subgroup.

The exercise is already solved since you know that there is only one 7-sylow, it has to be normal (here is a proof).

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Thank you very much. – Faisal Apr 6 '12 at 19:23

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