Clearly $G/H$ is cyclic if $G$ is cyclic. But sometimes $G/H$ is cyclic without $G$ necessarily cyclic, for example: $(\mathbb{Z}_6\times \mathbb{Z}_6)/\left \langle (2,3) \right \rangle \cong \mathbb{Z}_6$

In cases like the latter, how can one determine if the group is cyclic? I mean, is there something in general that lets me know?

For example

Write $(\mathbb{Z}_{20}\times \mathbb{Z}_6)/\left \langle (10,2) \right \rangle $ as an external direct product of cyclic groups of prime power order.

Is $(\mathbb{Z}_{20}\times \mathbb{Z}_6)/\left \langle (10,2) \right \rangle $ isomorphic to $\mathbb{Z}_{4}\times\mathbb{Z}_{5}$ or isomorphic to $\mathbb{Z}_{2}\times \mathbb{Z}_2\times \mathbb{Z}_5$?

  • 1
    $\begingroup$ Do you assume G to be abelian or not necessarily? $\endgroup$
    – Keen
    Mar 25 at 22:54
  • $\begingroup$ @Keen Well, I am working with Abelian groups, I missed $\endgroup$
    – Hopmaths
    Mar 25 at 22:55
  • $\begingroup$ You could try to calculate the order of $(0,1),(1,0)\in\mathbb Z_{20}\times\mathbb Z_6/\langle(10,2)\rangle$ $\endgroup$
    – Kenta S
    Mar 25 at 22:59
  • 2
    $\begingroup$ Smith normal form is often a useful tool for calculations of this type. In this case, you're looking at the cokernel of the matrix $\begin{bmatrix} 10 & 20 & 0 \\ 2 & 0 & 6 \end{bmatrix}$. $\endgroup$ Mar 25 at 23:58

According to the fundamental theorem of Abelian groups, you can write $G$ as a product $\mathbb{Z}_{n_1} \times \cdots \times \mathbb{Z}_{n_m}$. Now consider $f$ the morphism from $\mathbb{Z}^m$ to $G$, that consists of natural projections onto $n_i$. Let $\pi$ be the natural projection of $G$ onto $\frac{G}{H}$. Then $\frac{G}{H}$ is isomorphic to $\frac{\mathbb{Z}^m}{ker(\pi \circ f)}$.

Now we use the structure theorem of finitely generated modules over a PID. There exists a basis $(e_1 ,\cdots ,e_m)$ of $\mathbb{Z}^m$ and unique sequence of positive $d_1$ dividing $d_2$ dividing $d_3$ etcetera dividing $d_r$, such that $(d_1 e_1 , \cdots ,d_r e_r)$ is a basis of $ker(\pi \circ f)$, with $r$ being the rank of $ker(\pi \circ f)$. From this, we can simply deduce that $\frac{\mathbb{Z}^m}{ker(\pi \circ f)}$ is cyclic if and only if: $m=r$ and $d_1=\cdots = d_{m-1}=1$.

These $d_i$ can be calculated algorithmically if one knows a generating family of $ker(\pi \circ f)$. Now if we take $u_1, \cdots , u_k$ a family in $G$ generating $H$ (which is how one would typically describe a subgroup). Then if you take $v_1, \cdots , v_k \in \mathbb{Z}^m$, such that $f(v_i)=u_i$, then $v_1 , \cdots, v_m$ together with a family generating $ker(f)$ generate $ker(\pi \circ f)$.

Using that, you calculate the invariants $d_1, \cdots, d_m$ for the submodule $ker(\pi \circ f)$ and you can then conclude on whether $\frac{G}{H}$ is cyclic.

Now I am sure that for most cases you'll encounter, this method is an overkill, but it should work in general.


If $|G/H|$ is prime, then of course it's cyclic.

In general, the Smith normal form can be used to compute the structure of quotients of $\mathbb Z^n $. But every finitely generated abelian group is such a quotient. After determining $G/H$ as a quotient of the free abelian group on $n$ generators, this gives an approach.


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