Let $G$ be any group, and let $Z$ be its center.
(a) Show that $G/Z\cong \text{Inn}(G)$.
(b) Conclude that $\text{Inn}(G)$ cannot be a nontrivial cyclic group.
I've already gotten part (a) by considering the mapping $\pi:G\rightarrow\text{Inn}(G)$ such that $\pi(g)$ is the automorphism that takes $x$ to $g^{-1}xg$ for all $x\in G$. The mapping $\pi$ is clearly a surjective homomorphism with kernel $Z$, and part (a) follows from the isomorphism theorem.
For part (b), I must prove that $G/Z$ cannot be a nontrivial cyclic group. If it were, the group would equal $\{Z,Zg,Zg^2,\ldots,Zg^{n-1}\}$ for some $g\in G$. Also, $G/Z$ would be an abelian group, and it follows that the commutator subgroup $G'$ belongs to $Z$. I don't see how to derive a contradiction from there.
Best Answer
From $G/Z$ cyclic you can get something stronger: Let $a,b \in G$, then $a = z_1g^n$ and $b = z_2g^m$ for some $z_1,z_2 \in Z$. Now $ab = z_1g^n z_2g^m = z_2g^mz_1g^n = ba$. Thus $G$ is abelian, therefore $G = Z$. What now?