MHB Solution: Is the derived subgroup of a cyclic group always trivial?

[Chris L T521]

Here's this week's problem.

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Problem: Let $G$ be a group such that $G/Z(G)$ is cyclic. Show that the derived subgroup (commutator subgroup) of $G$ is trivial.

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[Chris L T521]

No one answered this week's question. You can find my solution below:

[sp]Proof: This statement is equivalent to proving that $G$ is Abelian. Suppose that $G/Z(G)$ is cyclic. Then there exists some generator $gZ(G)$ such that $<gZ(G)> = G/Z(G)$. Let $a,b\in G$ be arbitrary. Then for some $i$, $aZ(G)=(gZ(G))^i= g^i Z(G)$ and for some $j$, $bZ(G)=(gZ(G))^j=g^jZ(G)$. Now, $aZ(G) = g^iZ(G)\implies a=g^iz_1$ for some $z_1\in Z(G)$ and similarly $bZ(G) = g^jZ(G) \implies b=g^j z_2$ for some $z_2\in Z(G)$. Thus $ab=(g^iz_1)(g^jz_2) = g^ig^jz_1z_2$ since elements of the center comute with all elements of $G$. Therefore, we have
\[ab=g^ig^jz_1z_2 = g^{i+j}z_1z_2 = g^{j+i}z_2z_1 = g^jg^iz_2z_1 = g^jz_2 g^iz_1 = ba.\]
Since $a$ and $b$ were arbitrary, it follows that $G$ is Abelian. Hence $C=[G,G] = \{e\}$.$\hspace{.25in}\blacksquare$[/sp]