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This is Hungerford's Theorem II.1.2

Any two bases of a free abelian group $F$ have the same cardinality.

In his proof, he first assumes $F$ has a basis $X$ of finite cardinality $n$. Then he establishes $F/2F\simeq \bigoplus_1^n \mathbb{Z}_2$ and $|F/2F|=2^n$. Afterwards he considers another basis $Y$. Here's what I don't understand:

(quote) If $Y$ is another basis of $F$ and $r$ is any integer s.t. $|Y|\geq r$, then a similar argument shows that $F/2F|\geq 2^r$, whence $2^r\leq 2^n$ and $r\leq n$. It follows that $|Y|=m\leq n$ and $|F/2F|=2^m$. Therefore $2^m=2^n$ and $|X|=n=m=|Y|$.

I also don't understand his next line

(quote) If one basis of $F$ is infinite, then all bases are infinite by the previous paragraph.

Question. I assume his intention is to avoid claiming $F/2F$ is a $\mathbb{Z}_2$ vector space, because the book hasn't mentioned such a concept at this point yet. But even if so,

  1. he can simply have $|F/2F|=2^{|X|}$ and $|F/2F|=2^{|Y|}$, which can give $|X|=|Y|$ with $X$ assumed to be finite. Why the trouble of introducing integer $r$?
  2. if he considers $Y$ but not assuming it to be finite, that is, if $Y$ is infinite, then any integer $r$ is strictly smaller than $|Y|$. In this case, is his reasoning still valid?
  3. In the 2nd quote above, why is his claim true? That is, I assume he means his "previous paragraph" shows "if one basis is finite, then all bases are finite and have the same size", so if one basis if infinite, no other basis can be finite. But this goes back the question 2.

Hope it makes sense. Thanks in advance.

Update. It seems, if firstly consider a basis $Y$ without assuming finiteness, and get $|F/2F|\geq 2^r$ for every positive integer $r\leq |Y|$, then assuming existence of another finite basis $X$ and repeat his steps, no confusion would arise.

RobPratt
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user760
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    The first confusion should be cleared if you realize that the cardinality of $F/2F$ is an invariant of $F$, having nothing to do with its basis. – Absurdus Jun 11 '25 at 01:24
  • I see no point in claiming $F/2F$ a $\mathbb{F}_2$ vector space. It is a finite abelian group, and that's enough. – Absurdus Jun 11 '25 at 01:28
  • And how could there be finite and infinite bases at the same time? If you have a finite basis, then $F/2F$ is a finite group; If you have an infinite basis, then $F/2F$ is a infinite group. As for his $r$, I have no idea. I think the only trouble here is, if $X$ and $Y$ are both infinite bases, does the invariance of $|F/2F|$ imply that $|X| = |Y|$? This becomes a subject in set theory, rather than arithmetic; and I haven’t thought of an elementary proof yet. – Absurdus Jun 11 '25 at 02:26
  • @Absurdus, For $\mathcal{P}(X)$ and $\mathcal{P}(Y)$ equipotent implies $X$ and $Y$ are equipotent, you may take a look at https://math.stackexchange.com/questions/29366/do-sets-whose-power-sets-have-the-same-cardinality-have-the-same-cardinality – Krish Jun 11 '25 at 12:26
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    @Krish Thank you for your information. Set theorists never let me down on producing theorems against intuition. Nevertheless, after some thoughts, I think infinite direct sum is distinct from infinite direct product. Direct sum demands finite linear combination. The latter has cardinality $|\mathcal{P}(X)|$ as the former's is $|X|$. – Absurdus Jun 12 '25 at 02:42
  • @Absurdus You are right. $|F/2F|$ is not $2^{|X|}$ if $|X|$ is not finite. I didn't realize that. – user760 Jun 12 '25 at 14:42

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