Cardinality of quotient group

Question

I have a question.

$G$ is a free abelian group with rank $m$. $A \subset B$ are both subgroups of $G$ with rank $m$, but $A \neq B$.

Why is the inequality $|G /B| < |G /A|$ correct?

Thank you for the time and effort!

Answer 1

Hints:

1. Prove that if $G$ is free abelian of rank $m$ and $A$ is a rank $m$ subgroup of $G$ then $G/A$ is finite.

2. Prove that if $A\lneq B$ then $\phi: G/A\twoheadrightarrow G/B$ in a natural way, and $\ker\phi\neq1$.

3. Combine 1. and 2. to prove your result.

Answer 2

Define $\;\phi: G/A\to G/B\;,\;\;\phi gA:=gB\;$

The above is well defined since $\;gA=hA\implies h^{-1}g\in A\subset B\implies gB=hB\;$ , it's clearly a homomorphism of groups which is also trivially an epimorphism.

Choose now $\;b\in B\setminus A\;$ . Then $\;bA\neq A\;$, but $\;bB=B\;$ , and this implies $\;\phi\;$ cannot be injective (since $\;\ker\phi\neq1\;$ ...) . Finitiness of the quotient groups finishes now the proof.