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Prove that no group of order $56$ can be simple using steps

●finding sylows number 2-subgroups and sylow 7-subgroups
●explain why any of sylow 7-subgroups must intersect trivially, but this is not the case for sylow 2-subgroup
● use counting argument to show that if there is more than one sylow 7 subgroup, then there can be only one sylow 2 subgroup
● Show that GROUP of order $56$ is not simple.

my question is why does this argument fails to show that no group of order 112 can be simple

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  • $\begingroup$ Why do you expect the argument to fail? $\endgroup$ – Tobias Kildetoft Dec 12 '14 at 12:22
  • $\begingroup$ I think because the intersection.but I'm not sure $\endgroup$ – M Alrantisi Dec 12 '14 at 12:25
  • $\begingroup$ I mean, did you try to work through the same method and something failed? $\endgroup$ – Tobias Kildetoft Dec 12 '14 at 12:25
  • $\begingroup$ There exists a group of order $112$ with $|{\rm Syl}_7(G)|=8$ and $|{\rm Syl}_2(G)|>1$, so that part of the argument must fail! $\endgroup$ – Derek Holt Dec 12 '14 at 12:27
  • $\begingroup$ Yes ..sylow group number is not same $\endgroup$ – M Alrantisi Dec 12 '14 at 12:27
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If in a group of order $56 = 2^3 \cdot 7$ we have $8$ Sylow $7$-subgroups, then we get $8 \cdot (7-1) = 48$ elements of order $7$ (as the intersections of Sylow $7$'s are trivial). Thus we have $56 - 48 = 8$ elements left for the Sylow $2$-subgroups, which have order $8$. Thus there must be just one.

In a group of order $112 = 2^4 \cdot 7$ we cannot obtain the same by counting elements. It's possible that $N(7) = 8$ and $N(2^4) = 7$.

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  • $\begingroup$ Thanks it seems they different by counting theorem $\endgroup$ – M Alrantisi Dec 15 '14 at 19:55

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