Prove $G$ is Abelian if $N$ is in the centre of $G$ and $G/N$ is cyclic

Question

I need some help on this one.

$G$ is a group. If $N$ is a subgroup of $G$ contained in the centre of $G$ and $G/N$ is cyclic, show that $G$ is Abelian.

My attempt is only half way and stuck at the bit at the end...

If $G$ can be shown to be cyclic, then it is Abelian, since $G=<g>$ then $g_1,g_2 \in G$ are $g_1=g^n,g_2=g^m$ for some $n,m$ and $g_1g_2=g^ng^m=g^{n+m}=g^{m+n}=g_2g_1$.

By definition of a centre of $G$, $\forall n \in N$, $gn=ng$ for any element in $G$. Namely, $N$ must be a normal subgroup. (Here, I pondered if I should just proceed or change gears to looking at homomorphisms of $G$ since $N$ is normal).

Now $G/N$ is cyclic so $G/N=<gN>$ for some $g \in G$. Then, $\forall g'N \in G/N$ there is some $m$ so that $(gN)^m=g^mN=g'N$. Therefore, for the representative elements of $G/N$, I have $g^m$.

Here I get stuck. So if $g'N$ then I have that each $g'$ is cyclic but as an equivalent class, I cannot guarantee that all $g'$ can be represented in the form $g^m$. The only thing I can guess is the canonical map $p:G \to G/N$ that might be relevant to solving this but again, I don't know how specifically it would help.

Does anyone know how to solve this?

Answer 1

Your approach is doomed to failure, because you're trying to show that the group $G$ is cyclic, and this does not follow from the hypotheses. Indeed, take $G=C_2\times C_2$; then we have a normal subgroup $C_2\subset G$ that is contained in the centre of $G$ (since $G$ is abelian) and $G/C_2=C_2$ is cyclic. But $G$ is not cyclic.

I'm not going to tell you how to do this question, since it'll be better for you if you work it out on your own, but try to keep this example - and other similar ones - in mind when you investigate the question.

Answer 2

Hint: don't try showing $G$ is cyclic. Rather, suppose $gN$ is a generator for $G/N$, so for every $x\in G$ there exists $n$ such that $xN=g^mN$, that is, there exist $m$ and $x'\in N$ with $x=g^nx'$. So, if $x,y\in G$, you can write $x=g^mx'$ and $y=g^ny'$. Consider $xy$ and $yx$…

Answer 3

$G/Z(G) \cong (G/N) / (Z(G)/N) $ is a homomorphic image of the cyclic group $G/N$ and so is cyclic.

It is well known that if $G/Z(G)$ is cyclic then $G$ is abelian.

Answer 4

$(1) \ \langle xN \rangle = G/N$ for some $x \in G$.

$(2)$ Let $aN, bN \in G/N$ then $aN = (xN)^j, bN = (xN)^k$

$(3)$ Since $N$ is normal $(xN)^j = x^jN$ and $(xN)^k = x^kN$

$(4)$ Now $aN = x^jN \Rightarrow a = x^jn_1$ anf $bN = x^kN \Rightarrow b = x^kn_2$ for some $n_1, n_2 \in N$

$(5)$ Now show $ab = ba$ using these representations and the fact that $n_1, n_2$ commute

Answer 5

$G/N$ cyclic means everything in $G$ may be written in the form $g^m h$ for some $h\in N$. Combine this with the fact $N$ is contained in the centre to show $(g^a x)(g^b y) = (g^b y)(g^a x)$.