# $\text{Inn}(G)$ cannot be nontrivial cyclic group

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.

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?
$$\forall x,y\in G\;\exists z_1,z_2\in Z(G)\;,\;n_1,n_2\in\Bbb N\;\;s.t.\;\; x=g^{n_1}z_1\;,\;y=g^{n_2}z_2\implies$$
$$xy=g^{n_1}z_1g^{n_2}z_2=g^{n_1}g^{n_2}z_1z_2=g^{n_2}g^{n_1}z_2z_1=\ldots$$
The above thus proves $\,G\,$ is abelian, but then $\,Z(G)=G\,$ , so$\;\ldots\;$
• So $G/Z$ is the trivial group! May 26, 2013 at 18:19