This is a little bit of a tongue twisting, for want of a better. $G$ abelian is equivalent to $G/Z(G)$ cyclic.

Question

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Problem Restatement

We are asked to prove the following statement about a group (G):

Claim: (G) is abelian iff the quotient group (G/Z(G)) is cyclic.

Recall that

(Z(G)={\,z\in G\mid zg=gz\text{ for all }g\in G\,}) is the center of (G).
A group is abelian when every pair of elements commutes: (ab=ba) for all (a,b\in G).
A group is cyclic when it is generated by a single element.

So we must prove two directions:

If (G) is abelian, then (G/Z(G)) is cyclic.
If (G/Z(G)) is cyclic, then (G) is abelian.

Proof

1. (G) abelian (\Longrightarrow) (G/Z(G)) cyclic

Assume (G) is abelian.

Step 1 – Identify the center.
Because every element commutes with every other element, each element belongs to the center. Hence
[ Z(G)=G. ]

Step 2 – Form the quotient.
The quotient (G/Z(G)) is then (G/G). By definition of a quotient, the only coset is the trivial coset ({G}).

Step 3 – Recognize the structure.
A group that consists of a single element is the trivial group. The trivial group is cyclic (it is generated by its sole element).

Thus, when (G) is abelian, (G/Z(G)) is the trivial (hence cyclic) group.

[ \boxed{\text{If }G\text{ is abelian, then }G/Z(G)\text{ is cyclic.}} ]

2. (G/Z(G)) cyclic (\Longrightarrow) (G) abelian

Assume now that the quotient (G/Z(G)) is cyclic. Let us denote a generator of this cyclic group by the coset (gZ(G)) for some (g\in G). Hence

[ G/Z(G)=\langle gZ(G)\rangle . ]

Step 1 – Describe an arbitrary element of (G).
Every coset of (Z(G)) can be written as a power of the generator: [ xZ(G)= (gZ(G))^{n}=g^{\,n}Z(G) \quad\text{for some integer } n . ] Thus for any element (x\in G) there exists an integer (n) and a central element (z_{x}\in Z(G)) such that
[ x = g^{\,n}z_{x}. ] Similarly, for any (y\in G) there are (m\in\mathbb Z) and (z_{y}\in Z(G)) with
[ y = g^{\,m}z_{y}. ]

Step 2 – Compute the product (xy).
[ xy = (g^{\,n}z_{x})(g^{\,m}z_{y}) = g^{\,n}g^{\,m}z_{x}z_{y} = g^{\,n+m}z_{x}z_{y}. ]

Step 3 – Compute the product (yx).
[ yx = (g^{\,m}z_{y})(g^{\,n}z_{x}) = g^{\,m}g^{\,n}z_{y}z_{x} = g^{\,m+n}z_{y}z_{x}. ]

Step 4 – Use commutativity of central elements.
Since (z_{x},z_{y}\in Z(G)), they commute with each other and with every element of (G). In particular, [ z_{x}z_{y}=z_{y}z_{x},\qquad\text{and}\qquad g^{\,n+m}=g^{\,m+n}. ]

Therefore [ xy = g^{\,n+m}z_{x}z_{y}=g^{\,m+n}z_{y}z_{x}=yx . ]

Step 5 – Conclude.
We have shown that any two elements (x,y\in G) satisfy (xy=yx). Hence (G) is abelian.

[ \boxed{\text{If }G/Z(G)\text{ is cyclic, then }G\text{ is abelian.}} ]

3. Final Statement

Combining the two directions, we have proved the equivalence:

[ \boxed{\; G \text{ is abelian } \iff G/Z(G) \text{ is cyclic. \;} ]

The forward direction yields the trivial cyclic group, while the reverse direction forces every element of (G) to be a product of a power of a fixed element and a central element, which forces commutativity throughout the group.

Common Mistakes

Mistake	Why it’s wrong	How to avoid it
Assuming “cyclic” means “generated by a non‑identity element.”	The trivial group ({e}) is cyclic (generated by its only element). Ignoring this leads to thinking the forward direction fails.	Remember the definition: a group is cyclic if there exists an element (a) (possibly the identity) such that (\langle a\rangle = G).
Confusing the generator of (G/Z(G)) with an element that generates (G).	The element (g) that generates the quotient need not generate the whole group (G).	Work with cosets: write every element of (G) as (g^{n}z) with (z\in Z(G)).
*Forgetting that elements of the center commute with all* elements, not just with each other.**	If you only use (z_{x}z_{y}=z_{y}z_{x}) you miss the crucial step that (z) commutes with the powers of (g).	Explicitly note that (z\in Z(G)) implies (zg = gz) for any (g\in G); this justifies moving central factors past the powers of (g).
Treating the exponent (n) (or (m)) as a non‑negative integer only.	In a quotient group, the generator may have infinite order, so negative powers are also needed.	State that (n,m\in\mathbb Z) (allowing negative exponents) when writing (g^{n}).
Assuming the converse is “obvious” and skipping the detailed computation.	The implication (G/Z(G)) cyclic (\Rightarrow) (G) abelian is not trivial; a step‑by‑step verification is required.	Write out the expressions for arbitrary elements, multiply them in both orders, and use centrality to show equality.

Keeping these points in mind will help avoid pitfalls when dealing with similar problems about centers and quotient groups.

Original question: This is a little bit of a tongue twisting, for want of a better. $G$ abelian is equivalent to $G/Z(G)$ cyclic. on Mathematics Stack Exchange, licensed CC BY-SA.