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Let G be a group of order 56. Show that G is not simple.
We will use Sylow’s Theorem to show that either the 2-Sylow subgroup or 7-Sylow subgroup is normal.
By Sylow’s Theorem . Thus .
Also, . Therefore .
If or , we are done, as one of the Sylow subgroups is normal.
Suppose to the contrary and .
Number of elements of order 7 = 8 x (7-1)=48
Remaining elements = 56-48=8. This is just enough for one 2-Sylow subgroup, thus . This is a contradiction.
Therefore, a group of order 56 is simple.