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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.

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2 Answers 2

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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?

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Hints:

$$\forall x,y\in G\;\exists z_1,z_2\in Z(G)\;,\;n_1,n_2\in\Bbb N\;\;s.t.\;\; x=g^{n_1}z_1\;,\;y=g^{n_2}z_2\implies$$

$$xy=g^{n_1}z_1g^{n_2}z_2=g^{n_1}g^{n_2}z_1z_2=g^{n_2}g^{n_1}z_2z_1=\ldots$$

The above thus proves $\,G\,$ is abelian, but then $\,Z(G)=G\,$ , so$\;\ldots\;$

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  • $\begingroup$ So $G/Z$ is the trivial group! $\endgroup$
    – Paul S.
    May 26, 2013 at 18:19
  • $\begingroup$ Indeed so....you can harvest the contradiction from wherever you want. $\endgroup$
    – DonAntonio
    May 26, 2013 at 19:53

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