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.