# When $G/H$ is cyclic?

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 $$\mathbb{Z}_{2}\times \mathbb{Z}_2\times \mathbb{Z}_5$$?

• Do you assume G to be abelian or not necessarily?
– Keen
Mar 25 at 22:54
• @Keen Well, I am working with Abelian groups, I missed Mar 25 at 22:55
• You could try to calculate the order of $(0,1),(1,0)\in\mathbb Z_{20}\times\mathbb Z_6/\langle(10,2)\rangle$ Mar 25 at 22:59
• 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 $\begin{bmatrix} 10 & 20 & 0 \\ 2 & 0 & 6 \end{bmatrix}$. 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.