Let $L$ be a subgroup of $\mathbb{Z}^3$ of index $16$. What are the possibilities for $\mathbb{Z}^3 /L$?

Question

Let $L$ be a subgroup of $\mathbb{Z}^3$ of index $16$. What are the possibilities for $\mathbb{Z}^3 /L$?

Since $L$ has 16 elements, I think it might be $\mathbb{Z}_{16},\mathbb{Z}_2 \oplus \mathbb{Z}_8, \mathbb{Z}_4 \oplus \mathbb{Z}_4,$ or $\mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_4$. But I don't know how to get the possible results of $\mathbb{Z}^3/L$ So please help me to figure it out. Thank you!

Answer 1

If $L\subset\Bbb{Z}^3$ is a subgroup of index $16$ then its quotient $\Bbb{Z}^3/L$ is a group of order $16$. Any such subgroup $L\subset\Bbb{Z}^3$ is normal because $\Bbb{Z}^3$ is abelian, so this quotient always exists.

Because $\Bbb{Z}^3$ is abelian, also every quotient of it is an abelian group. Moreover, because $\Bbb{Z}^3$ is generated by three elements, so is every quotient of it. Hence all such quotients are abelian groups of order $16$ generated by three elements.

Conversely, if $G$ is an abelian group of order $16$ generated by three elements, the map $$\Bbb{Z}^3\ \longrightarrow\ G,$$ mapping three generators of $\Bbb{Z}^3$ to three generators of $G$, is surjective, so $G$ is isomorphic to a quotient of $\Bbb{Z}^3$ by a subgroup of index $16$. This shows that all abelian groups of order $16$ generated by three elements arise a quotients of $\Bbb{Z}^3$ by a subgroup of index $16$.

By the fundamental theorem on finite abelian groups, the abelian groups of order $16$ are precisely $$\Bbb{Z}_{16},\qquad\Bbb{Z}_8\oplus\Bbb{Z}_2,\qquad\Bbb{Z}_4\oplus\Bbb{Z}_4,\qquad\Bbb{Z}_4\oplus\Bbb{Z}_2\oplus\Bbb{Z}_2,\qquad\Bbb{Z}_2\oplus\Bbb{Z}_2\oplus\Bbb{Z}_2\oplus\Bbb{Z}_2,$$ where only the last one is not generated by three elements. This gives us all four possibilities for $\Bbb{Z}^3/L$, up to isomorphism.

Answer 2

Being the quotient of a (free) abelian group on 3 generators it has to be an abelian group of order 16 generated by 3 or less elements.

This boils down to factorizing 16 into 3 or less number of positive integers. TA factorization $16=abc$ is to be interpreted as the abelian group of order 16 obtained as the direct product of 3 cyclic groups having orders $a,b,c$ respectively. ($1\le a\le b\le c\le 16)$. (ordering on a,b,c is needed to avoid isomorphic duplication).

Not all abelian groups of order 16 are obtainable this way: for example two copies of Klein's group $V_4\times V_4$ is not possible,