Let a, b≠5, then a*b≠5.
②(a*b)*c=[ab-5(a+b)+30]*c
=[a B- 5(a+b)+30]c-5[a B- 5(a+b)+30+c]+30
= ABC-5(a b+ BC+ca)+25(a+b+ c)- 120,
Similarly, a*(b*c)=a*[bc-5(b+c)+30]
= a[BC-5(b+c)+30]-5[a+BC-5(b+c)+30]+30
= ABC-5(a b+ BC+ca)+25(a+b+ c)- 120,
So (a*b)*c=a*(b*c)
③ let a*x=ax-5(a+x)+30=a,
ax-5a-5x+30=a,
(a-5)x=6(a-5),
a≠5,
So x=6, that is, 6 is the unit element of r'.
④ let a*y=ay-5(a+y)+30=6,
(a-5)y=5a-24,
Y=(5a-24)/(a-5), obviously not 5,
So the reciprocal of a is (5a-24)/(a-5).
All in all,
Yi Zhi a*b=b*a,
So < r', *> is an exchange group.
4*4=(4-5)(4-5)+5=6,
So |4|=2.