Group of order 56 is not simple + Affordable Air Purifier

Singapore Maths Tuition

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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.

$latex |G|=2^3cdot 7$

By Sylow’s Theorem $latex n_2mid 7, n_2equiv 1pmod 2$. Thus $latex n_2=1,7$.

Also, $latex n_7mid 8, n_7equiv 1pmod 7$. Therefore $latex n_7=1, 8$.

If $latex n_2=1$ or $latex n_7=1$, we are done…

View original post 54 more words

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