Question

Prove that G is an abelian group if and only if$\mathbf{\text{Inn}}\mathit{G}$ consists of a single element.

Short answer

Itproved that G is an abelian group if and only if$\text{Inn}G$ consists of a single element.

Step-by-step solution

Step-by-Step Solution Step 1: Condition for abelian and non-abelian group

Group $G$is said to be abelian if $ab=ba$for all$a,b\in \mathbf{G}$, and group$G$is said to be non-abelian if$ab\cancel{=}ba$for all$a,b\in \mathbf{G}$.

Prove that G is an abelian group

Assume that $G$is abelian .

Take an arbitrary element in $f\left(a\right)=c^{-1}ac\in \text{Inn}G$because localid="1651573735341" $G$is abelian.

Then,

$$\begin{gathered} f\left(a\right)=c^{-1}ac \\ =a{c}^{-1}c \\ =a \end{gathered}$$

Therefore, f is an identity function.

Assume that localid="1651573913661" $\text{Inn}G$contains only one element localid="1651574258366" $f\left(a\right)=b^{-1}ab$for some localid="1651573901386" $b\in G$.Then, it will be true that localid="1651573947113" $b^{-1}ab=b^{-1}{c}^{-1}acb$for all a in localid="1651574002956" $G$.Now, localid="1651574038904" $c^{-1}ac=a$and consequently,

$ac=ca\left(\text{for\hspace{0.17em}all\hspace{0.17em}}a\text{\hspace{0.17em}in\hspace{0.17em}}G\right)$

Since localid="1651574264200" $c$commutes with all elements in localid="1651574278025" $G$andlocalid="1651574269056" $c$was arbitrary. Therefore, it is proved that$G$is abelian.