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I've started reading Pinter's Book of Abstract Algebra, and one of the early exercises calls for a proof that in a finite abelian group $G$, $(a_1a_2a_3a_4...a_n)^2 = e$, if there are any $a_n$ that are their own inverses, and that if no $a_n$ is its own inverse, then $(a_1a_2a_3a_4...a_n) = e$.

Since I'm not doing this for a class, I worked out a very informal proof for the second case (no $a_n$ is its own inverse): Since there must be an $a^{-1}$ for each $a$, then $(a_1a_2a_3a_4...a_n)$ can be reduced to so many instances of $(a_1a_1^{-1}a_2a_2^{-1}...a_{n/2}a_{n/2}^{-1})$, which can be simplified to $e^{n/2}$, or just $e$.

Is there a particular reason why this holds only for finite abelian groups?

If one were to list rational numbers in order, as Cantor demonstrated, would the product of this list tend toward one?

Also, would it be fair--albeit relatively meaningless--to say that the group $\mathbb{Q} > 0$ has an odd number of members, as it contains exactly one member that is its own inverse?

(full disclosure--this is basically a hobby for me, so apologies in advance for any obvious errors in logic/notation/etc., but please point them out)

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    $\begingroup$ First off, you are going to run into trouble defining the product in the first place. How do infinite products work? When do they converge? Will your product converge? Second, there is an issue related to Riemann's rearrangement theorem that you have to consider; namely, order matters. $\endgroup$ – Xander Henderson Feb 18 '18 at 17:04
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    $\begingroup$ As for the parity of the number of elements of $\mathbb{Q}$, there is a one-to-one correspondence between $\mathbb{Q}$ and $\mathbb{Q}\setminus\{1\}$. If the first set has odd parity, then the second must, too, no? So that seems like a fairly meaningless statement. $\endgroup$ – Xander Henderson Feb 18 '18 at 17:05
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    $\begingroup$ Good question. "Is there a particular reason why this holds only for finite abelian groups? " Yes, because they are finite. You can't shunt things indefinately so they don't work. This is analagous to "the sum of all integers is zero" because $0+1+(-1) + 2+(-2)+...$. But we can just as easily rearrange them as $0 + 1 + (2-1) + (3-2) + (4-3) +...$ to get that the sum is equal to $1$. etc. $\endgroup$ – fleablood Feb 18 '18 at 17:08
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    $\begingroup$ You can probably do the exact same argument and reach the conclusion that the product of all rational numbers is $2$. $\endgroup$ – Arthur Feb 18 '18 at 17:09
  • $\begingroup$ @XanderHenderson Riemann's rearrangement theorem! I forgot about that!! Could you post that as an answer and not as a comment? I'm not quite sure I follow your second comment. Can you elaborate? $\endgroup$ – Rich Jensen Feb 18 '18 at 17:10
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Infinite products are delicate. Generally, we say that an infinite product $$ \prod_{k=1}^{\infty} a_n $$ converges if and only if the series $$ \sum_{k=1}^{\infty} \log(a_n) $$ converges to some value $\log(a)$. In this case, the product converges to $a$. Otherwise, the product diverges.

To understand why this is a reasonable definition, recall that for two elements, we have $$ a_1 \cdot a_2 = a \iff \log(a_1) + \log(a_2) = \log(a); $$ and that for any finite product we can show by induction that $$ \prod_{k=1}^{n} a_k = a \iff \sum_{k=1}^{n} \log(a_k) = \log(a). $$ Thus we can understand the properties of products by reducing them to series using the logarithm (since you are reading about abstract algebra, you might consider how the multiplicative group of positive real numbers relates to the additive group of real numbers; are they isomorphic?).

Notice that there are several interesting ways for a product to diverge: it can oscillate (as a series might), it can diverge to infinity (consider the infinite product $\prod 2$), or it can diverge to zero! For example, if $\prod \frac{1}{2} = a$, then $$ \sum_{k=1}^{\infty} \log\left( \frac{1}{2} \right) = \log(a). $$ But the sum diverges to negative infinity, hence the product does not converge.

So, suppose that $\{q_k\}$ is an enumeration of the rational numbers. Then $$ \prod_{k=1}^{\infty} q_k = q \iff \sum_{k=1}^{\infty} \log(q_k) = \log(q). $$ There is no rearrangement of the rational numbers such that the series above converges. To see this, suppose that $\{q_n\}$ is an enumeration of the rationals. There are infinitely many $n \in \mathbb{N}$ such that $|\log(q_n)| > 1$, which implies that $|\log(q_n)|$ does not converge to zero as $n\to \infty$. Hence the series $\sum \log(q_n)$ does not converge (the general term doesn't go to zero).


As per Asaf Karagila's comments (below), a set is even if it can be split into two sets of the same "size", or if it can be split into pairs (this characterization requires the Axiom of Choice, but this is a technical point that is beyond the current scope). If a set cannot be split into pairs, then it is odd. Since any infinite set can be split into pairs (again, this requires choice), every infinite set—including the rationals—is even.

As such, this argument basically asserts that the concept of "parity" is more or less meaningless for infinite sets.

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  • $\begingroup$ Thanks for the clarification re: infinite cardinals. It would be interesting to find an arrangement of $\mathbb{Q}$ that converged, but that's probably well above my pay grade. $\endgroup$ – Rich Jensen Feb 18 '18 at 17:43
  • $\begingroup$ I think I see where you're going WRT Cauchy. Any series that converged would presumably have to oscillate around the limit (1), and in order for those oscillations to converge, one would need to start with, in layman's terms, the largest & smallest rational numbers, which, of course, cannot be defined. Correct? $\endgroup$ – Rich Jensen Feb 18 '18 at 18:12
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    $\begingroup$ @Javier Indeed, it does seem strange, it is the way that ultimately makes the most sense. The problem is that 0 is a branch point for the complex logarithm, and therefore causes all manner of badness when you try to say that an infinite product converges there. There are important reasons to insist that products cannot converge to zero. For example, working with things like the Euler product formula corresponding to a Dirichlet series. $\endgroup$ – Xander Henderson Feb 19 '18 at 3:31
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    $\begingroup$ @PeterLeFanuLumsdaine Well, can't I do that by grouping factors together as needed, just as I did in my answer? So e.g. I calculate something like $1\cdot(\frac{1}{2}\cdot\frac{2}{1})\cdot(\frac{1}{3}\cdot\frac{3}{1})\cdot(\frac{2}{3}\cdot\frac{3}{2})\cdot\ldots$? With logarithms, we would efficiently add up 0 infinite times, so it is well defined. This isn't thought as an counterexample to your statement, just saying that "multiplying all positive reals" doesn't assumes an order of associtivity. $\endgroup$ – SK19 Feb 19 '18 at 9:24
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    $\begingroup$ Yes, all infinite cardinals are even (again, assuming choice). I didn't claim that this makes the notion of "even" meaningful for infinite cardinals. Just that it exists... :) $\endgroup$ – Asaf Karagila Feb 19 '18 at 15:39
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In your proof you are making the assumption that $n$ is even and that the inverse of $a_k$ for $1\leq k\leq \frac{n}{2}$ is not in $\{a_1,\ldots,a_{\frac{n}{2}}\}$, so in general, your idea doesn't work.

Also your theorems are flawed. They should read like this:

  1. For any finite group $G$ with elements $a_1,\ldots,a_n$ holds $(a_1\ldots a_n)^2=e$.
  2. For any finite group $G$ with elements $a_1,\ldots,a_n$ such that the only selfinverse element is $e$ holds $a_1\ldots a_n=e$.

The proof of (2.) is simple. Because of (1.) we know that the element of $G$ of form $a_1\ldots a_n$ is selfinverse (because $(a_1\ldots a_n)^2=e$), but with our assumption in (2.) we then get $a_1\ldots a_n=e$ immediatly.

As for your question: You have the problem that it is not well defined in which order your elements are multiplied. Sure, you can have an order such that the product is $1$. But you could also have an order such that the product is $\frac{1}{2}$ or $2$ or any other number. Even $0$ and $\infty$ are possible. The terms "convergence" and "absolute convergence" are needed here. See for example wikipedia (there it is about sums, but it is the same principle).

For illustration, your claim of the product being one is roughly equivalent to saying that the sum of all integers is $0$. I'm using that because listing all rationals is always a bit tricky.

Lets make an attempt:

$$0+1+(-1)+2+(-2)+\ldots=0+(1+(-1))+2+(-2)+\ldots=0+0+(2+(-2))+\ldots$$

Yeah, $0$ seems reasonable. But wait, what if we do it like this?

$$0+1+2+(-1)+3+4+(-2)+5+6+(-3)+\ldots=0+(1+2+(-1))+(3+4+(-2))+(5+6+(-3))+\ldots = 0+2+5+8+\ldots$$

Clearly we are still adding up all integers (in a way most people would probably not), but suddenly our sum gets bigger and bigger (forever).

Summation order (and multiplication order) often plays a critical role when dealing with infinite sums (or infinite products).

In conclusion you can't say "The product of all positive rationals is $1$", but "There is a way to multiply all positive rationals such that the product is $1$".

EDIT: For the people who are nagging about how using the associative law here makes my answer somehow not mathematically right, not in the spirit of the question or that I would just

write "1" in a countable number of funny ways

if I were to use this technique for multiplying all positive rationals to $1$, I'm proud to tell you that this technique is commonly used.

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  • $\begingroup$ @MichaelBurr And thanks to your comment I read it again and noticed a missing "positive", by which way I would have included 0 accidentally :) $\endgroup$ – SK19 Feb 18 '18 at 17:51
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    $\begingroup$ This seems like a better place to comment. When taking a sum, we cannot group like that. We take the limit of the partial sums. So $1 + -1 + 2 + -2 + \dots$ has partial sums of $1, 0, 2, 0, \dots$. The limit of the partial sums oscillates and it does not converge. $\endgroup$ – garyF Feb 19 '18 at 10:29
  • $\begingroup$ @garyF I'm not saying $\sum a_k$ is the same as $\sum (a_{2k-1}+a_{2k})$, it clearly isn't as I tried to point out in my answer. I'm saying the phrase "sum up an infinite set of numbers" does not indicate any order or associativity, so it is a valid way to answer the question with $\sum (a_{2k-1}+a_{2k})$ instead of $\sum a_k$ $\endgroup$ – SK19 Feb 19 '18 at 10:38
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    $\begingroup$ Anytime you see the word "clearly," you should be suspicious that something is being swept under the rug. I don't think that it is clear that any of your funny summation techniques actually represent "summing all of the integers." You are writing divergent sums with terms that involve all of the integers. It is not the same thing. $\endgroup$ – Xander Henderson Feb 19 '18 at 14:16
  • $\begingroup$ The document to which you have linked does not assert that you can multiply the rationals to get 1. It asserts that the harmonic series diverges. Also note that the argument does not devolve into clever associations, but is really more about how you organize the partial sums so that you can clearly see that the series diverges. That argument cannot be used to show that anything converges. $\endgroup$ – Xander Henderson Feb 19 '18 at 19:36
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Consider the sequence $\{s_n\}$ where $$ s_n=\begin{cases}n/2&n\text{ is even}\\ -(n-1)/2&n\text{ is odd} \end{cases}. $$ Consider the sum $$ \sum_{n=0}^\infty s_n. $$ This sum does not converge, even though every time $k$ is odd, $$ \sum_{n=0}^k s_n=0. $$

Suppose that we pick an order on the rationals greater than $1$, so $\{a_n\}$ is a sequence of all rationals greater than $1$. Then, you're considering the product $$ \prod_{n=0}^\infty a_na_n^{-1}=1. $$ It is true that this product is $1$ because every factor is $1$. However, if you were to look at the sequence $\{b_n\}$ where $$ b_n=\begin{cases} b_n=a_{n/2}&n\text{ is even}\\ b_n=a_{(n-1)/2}^{-1}&n\text{ is odd} \end{cases} $$ Then, the product $$ \prod_{n=0}^\infty b_n $$ has the same problem as before since it is $1$ whenever $n$ is odd, but can be any rational number greater than $1$ when $n$ is even.

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