Question

Question: (b) Prove that $\text{Inn}{D}_4\cong {\mathbb{Z}}_2\times {\mathbb{Z}}_2$

Short answer

It is verified that the group of inner automorphism of $D_4$is of order$4$.

Step-by-step solution

 Step 1: Inner Automorphism Groups

Suppose that G is a group. So, for a fixed $\mathit{a}\mathbf{\in }\mathit{G}$, the isomorphism of the form ${\mathit{f}}_{\mathbf{a}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathit{a}}^{\mathbf{-}\mathbf{1}}\mathit{x}\mathit{a}$is called an inner automorphism.

Prove that InnD4≅ℤ2×ℤ2

We need to prove that the group of inner automorphism of $D_4$is isomorphic to ${\mathbb{Z}}_2\times {\mathbb{Z}}_2$.

To prove the required condition, it is sufficient to show that every non-identity element of $\text{Inn}{D}_4$is of order .

Compute the order of the elements of $\text{Inn}{D}_4$.

$\begin{array}{c}{f}_{R_0}\circ f_{R_0}=f_{R_0{R}_0}=f_{R_0}\\ f_{R_{90}}\circ f_{R_{90}}=f_{R_{90}{R}_{90}}=f_{R_0}\\ f_h\circ f_h=f_{h^2}=f_{R_0}\\ f_d\circ f_d=f_{d^2}=f_{R_0}\end{array}$

This implies that,

$\begin{array}{c}\left|f_{R_0}\right|=1\\ \left|f_{R_{90}}\right|=2\\ \left|f_h\right|=2\\ \left|f_d\right|=2\end{array}$

Therefore, the group of inner automorphism of $D_4$is a non-cyclic group of order 4 and hence, it is isomorphic to${\mathbb{Z}}_2\times {\mathbb{Z}}_2$ .

Hence, it is proved that$\text{Inn}{D}_4\cong {\mathbb{Z}}_2\times {\mathbb{Z}}_2$ .