Group of order 56 is not simple + Affordable Air Purifier

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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.

|G|=2^3\cdot 7

By Sylow’s Theorem n_2\mid 7, n_2\equiv 1\pmod 2. Thus n_2=1,7.

Also, n_7\mid 8, n_7\equiv 1\pmod 7. Therefore n_7=1, 8.

If n_2=1 or n_7=1, we are done, as one of the Sylow subgroups is normal.

Suppose to the contrary n_2=7 and n_7=8.

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 n_2=1. This is a contradiction.

Therefore, a group of order 56 is simple.


About mathtuition88
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