Let G be a nonabelian group of order 10 having golden ratio 1+√
5
2
as the neutral
element. Then:
(a) Verify class equation for G.
(b) Find all elements of Inn(G).
(c) Verify that G/Z(G) ∼= Inn(G).
(d) For some elements x, y, z ∈ G: verify the commutator identity: [xy, z] = [x, z][x, z, y][y, z]
Part a.
Since the order of G is 10, so the golden ratio divides the order of G.
"=\\frac{|G|}{1+\\frac{\\sqrt{5}}{2}}\\\\\n=\\frac{|10|}{1+\\frac{\\sqrt{5}}{2}}\\\\\n=\\frac{2*10}{1+\\sqrt{5}}\\\\"
So, the class equation for G is
"[1+ \\frac{\\sqrt{5}}{2}]^n" where n = 1,...,10
Part b
In this part of the question, we will employ the general class equation for G.This will help us get the Inn(G) elements (First four elements).
The Inn(G) elements are given below
"[1+ \\frac{\\sqrt{5}}{2}]^n\\\\\n[1+ \\frac{\\sqrt{5}}{2}]^0= [1]\\\\\n[1+ \\frac{\\sqrt{5}}{2}]^1= [1+ \\frac{\\sqrt{5}}{2}]\\\\\n[1+ \\frac{\\sqrt{5}}{2}]^2=[1+ \\frac{\\sqrt{5}}{2}]^2\\\\\n[1+ \\frac{\\sqrt{5}}{2}]^3=[1+ \\frac{\\sqrt{5}}{2}]^3\\\\"
"[1],[1+ \n2\n5\n\u200b\n \n\u200b\n ],[1+ \n2\n5\n\u200b\n \n\u200b\n ] ^\n2\n ,[1+ \n2\n5\n\u200b\n \n\u200b\n ] ^\n3"
Part c
"=\\frac{G}{Z(G)}\\\\\n=\\frac{10}{2}\\\\\n=2"
and
"Inn(G)=\\frac{10}{2}=5\\\\\nSo,\\frac{G}{Z(G)}= Inn(G)."
Part d
From the question we are asked to verify the commutator identity.
"LHS\\\\\n[xy,z]=xyz-zxy\\\\\nRHS\\\\\n[x, z][x, z, y][y, z] = x[y,z]+[x,z]y\\\\\n\\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space = x[yz-zy]+[xz-zx]y\\\\\n\\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space = xyz-xzy+xzy-zxy\\\\\n\\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space \\space = xyz-zxy\\\\"
The LHS and the RHS are equal. Hence the commutator is identical.
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