Prove that $ x \in Z(G) \to G \ncong \text{Inn}(G) $. Let $G$ be a group and $x, a \in G$ where $x=a^2$. Then $x\in Z(G) G≇ Inn(G)$. I have already shown that if $x=b^{-1}a$, then $G≇Inn(G)$ since $\phi_{a}=\phi

Prove that $ x \in Z(G) \to G \ncong \text{Inn}(G) $. Let $G$ be a group and $x, a \in G$ where $x=a^2$. Then $x\in Z(G) G≇ Inn(G)$. I have already shown that if $x=b^{-1}a$, then $G≇Inn(G)$ since $\phi_{a}=\phi_{b}$. However, I don't think this argument applies for the cause that $x=a^2$. Does anyone have any hints on proving this using properties of isomorphisms?

_Hint._ Let $\phi:G\rightarrow \operatorname{Aut}(G)$ be defined by $\phi(g)=\theta_g$, where $\theta_g$ is the automorphism $\theta_g(x)=g^{-1}xg$.

1. What is the image of $G$ under $\phi$?

2. What is $\operatorname{ker}\phi$?

> Is $\phi$ an isomorphism?