# 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.

Proof:

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.

## Author: mathtuition88

https://mathtuition88.com/

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