Question

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]

Answer 1

Part a.

Since the order of G is 10, so the golden ratio divides the order of G.

$=\frac{|G|}{1+\frac{\sqrt{5}}{2}}\\ =\frac{|10|}{1+\frac{\sqrt{5}}{2}}\\ =\frac{2*10}{1+\sqrt{5}}\\$

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\\ [1+ \frac{\sqrt{5}}{2}]^0= [1]\\ [1+ \frac{\sqrt{5}}{2}]^1= [1+ \frac{\sqrt{5}}{2}]\\ [1+ \frac{\sqrt{5}}{2}]^2=[1+ \frac{\sqrt{5}}{2}]^2\\ [1+ \frac{\sqrt{5}}{2}]^3=[1+ \frac{\sqrt{5}}{2}]^3\\

$[1],[1+ 2 5 ​ ​ ],[1+ 2 5 ​ ​ ] ^ 2 ,[1+ 2 5 ​ ​ ] ^ 3$

Part c

$=\frac{G}{Z(G)}\\ =\frac{10}{2}\\ =2$

and

$Inn(G)=\frac{10}{2}=5\\ So,\frac{G}{Z(G)}= Inn(G).$

Part d

From the question we are asked to verify the commutator identity.

LHS\\ [xy,z]=xyz-zxy\\ RHS\\ [x, z][x, z, y][y, z] = x[y,z]+[x,z]y\\ \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\\ \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\\ \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.