December 22nd, 2017, 12:58 AM  #1 
Newbie Joined: Apr 2017 From: India Posts: 23 Thanks: 0  Abstract Algebra
Prove that if G is a group such that G/Z(G) is cyclic, then G is abelian. (I am unable to connect logic)I don't know why? 
December 22nd, 2017, 02:25 AM  #2 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 169 Thanks: 51 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Say the coset of $Z(G)$ represented by $g \in G$ generates $G/Z(G)$. Then every coset of $Z(G)$ is represented by $g^k$ for some $k$. Take any two elements $a, b$ of $G$. $a$ must be contained in a coset of $Z(G)$, say it's in the one represented by $g^n$. This means $a = g^n x$ for some $x \in Z(G)$. Similarly, $b = g^m y$ for some $m$ and some $y \in Z(G)$. Now it's straightforward to show that $a$ and $b$ commute.


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