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Let G be a group of order 56. Show that G is not simple.

Proof:

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.

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