3 years ago

Please help there is a attachment to this i think.

Mathematics

1 answer:

ser-zykov [4K]3 years ago

5 0

Answer:

squrt6(84)

Step-by-step explanation:

By the law:

squrtn(a^m) = a^(m/n)

Best regards

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$\Rightarrow$

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$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .