Can $G/2G$ and $\mathbb Z / 2\mathbb Z \times ... \times \mathbb Z / 2\mathbb Z$ be not isomorphic? $G$ is a free abelian group with finite basis.

Question

This is from Theorem 67.8 of Munkres Topology:

The proof gives stronger than bijective correspondences, specifically isomorphisms between $G$ and $\mathbb Z \times ... \times \mathbb Z$ and $2G$ and $2\mathbb Z \times ... \times 2\mathbb Z$. Either:

1. We also get an isomorphism between $G/2G$ and $\mathbb Z / 2\mathbb Z \times ... \times \mathbb Z / 2\mathbb Z$, but Munkres wanted to emphasize the bijection part of the isomorphism.

2. We don't necessarily get an isomorphism, but we necessarily get a bijective correspondence.

Which is it?

Answer 1

First and foremost: we do NOT get a bijection or isomorphism between $G/2G$ and $\mathbb{Z}/2\mathbb{Z}$: the group $G/2G$ is isomorphic to the product of $n$ copies of $\mathbb{Z}/2\mathbb{Z}$, so to the group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \ldots \times \mathbb{Z}/2\mathbb{Z}$, which is much bigger.

However, with this caveat the question is still interesting.

In the rest of the answer I will focus on the $n = 2$-case to simplify notation, the general case is similar.

I think we could reformulate your question as follows:

Given a group $Z$ and a normal subgroup $N$. It is easy to see that $N \times N$ is a normal subgroup of $Z \times Z$ and also that the quotient $(Z \times Z)/(N \times N)$ is in bijective correspondence with the product $(Z/N) \times (Z/N)$. Question: is it true or not that for all $Z$ and $N$ we actually have an isomorphism of groups between $(Z \times Z)/(N \times N)$ and $(Z/N) \times (Z/N)$?

(I hope you agree that the case $Z = \mathbb{Z}, N = 2\mathbb{Z}$ is the $n=2$-case of your question, with 'true' corresponding to option 1 and 'not true for all $Z, N$' corresponding to option 2)

The answer to the question is: YES. The sort-of-obvious isomorphism between sets which I figure you can find yourself is also a group homomorphism (which you can check from the definition once you have written down the set-isomorphism) and hence indeed an isomorphism of groups.

The intuition behind this is that the 'things' at the left and right hand sides of the $\times$ sign hardly interact with each other (even in the group case) so you can check everything you need to check 'one term at a time'.

Answer 2

You absolutely do get an isomorphism (you could try and prove that as an exercise). However, for the proof to work you only need that there is a bijection. So your alternative (1) holds.