Let g be a simple group of order 168. (a) show that if p is a sylow 7-subgroup of g then the normalizer ng(p) of p has order 21. (b) show that g contains no subgroup of order 14.

i think it's ф or x=6,-6

step-by-step explanation:

-x^2=-36

x^2=-36

ф

or

if you mean by : -(x^2)=-36

it'll be: x^2=36

x=6,x=-6

hope it'll ! have a good day!

you for asking!

your answer   is c

step-by-step explanation:

its less than 713.21 but greater than 5.2

a. true

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