If $\displaystyle \bigoplus_{i=1}^{n} \mathbb{Z} \cong \bigoplus_{i=1}^{m} \mathbb{Z}$ as groups, then $n=m.$

Question

If $\displaystyle \bigoplus_{i=1}^{n} \mathbb{Z} \cong \bigoplus_{i=1}^{m} \mathbb{Z}$ as groups, then $n=m.$

Here is my proof:

Let $G$ be a group such that $\varphi:G \rightarrow \displaystyle \bigoplus_{i=1}^{n} \mathbb{Z}$ is a isomorphism, consider the subgroup of $2G:=G+G$ then, define $2\varphi:2G \rightarrow \displaystyle \bigoplus_{i=1}^{n} \mathbb{2Z}$, such that $2\varphi(x):=\varphi (x)+\varphi (x), $ is a isomorphism, then $G/2G \cong \displaystyle \bigoplus_{i=1}^{n} \mathbb{Z_2}$, thus, $|G/2G|=2^n$.

Finally, if $\displaystyle \bigoplus_{i=1}^{n} \mathbb{Z} \cong G\cong \displaystyle \bigoplus_{i=1}^{m} \mathbb{Z}$.

$2^m=|G/2G|=2^n$, we conclude that $n=m$.

Is my proof ok?

Answer 1

The idea of your proof is fine, but there are some issues. (Contrary to the comments, it is not "OK".)

The main issue is with your function $2\varphi: 2G\rightarrow\bigoplus_{i=1}^{n} \mathbb{2Z}$, which you define by $2\varphi(x):= x+x$. Two issues:

1. The definition should be $2\varphi(x):= \varphi(x)+\varphi(x)$, as otherwise your codomain is incorrect ($x\in 2G$ not $\bigoplus_{i=1}^{n} \mathbb{2Z}$).
2. This is not the function you want. This function "doubles" elements, and you have already doubled every element by taking $2G$. In doubling them again you are essentially taking $4G$, so your quotient group, roughly, would be $$G/4G \cong \displaystyle \bigoplus_{i=1}^{n} \mathbb{Z_4}.$$ The function you are wanting is the restriction of $\varphi$ to $2G$, which is written $\varphi|_{2G}$. If you had used this function instead then everything else would have been fine.

Three other, more minor comments:

• As @GregMartin commented, $2G\neq G+G$. Instead, $2G=\{x+x\mid x\in G\}$.
• Maybe you could have justified the claim that $G/2G \cong\bigoplus_{i=1}^{n} \mathbb{Z_2}$? It is true, but if this was an exam the marking scheme might want you to prove this.
• Don't use \displaystyle in your MathJax, as it can look weird. Instead, either put it on its own line, via $$...$$, or leave it as it is. For example, in the above I have used $\bigoplus_{i=1}^{n}$ "inline", but if I had put a \displaystyle infront, e.g. $\displaystyle\bigoplus_{i=1}^{n}$, it is distracting because it messes with spacing between lines (I've hopefully made my point but and just typing to add text to emphasise the line spacing issue).