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Let $ (G,\ast)$ be a group with identity $e$ and cardinality $2n$ for some $n\in\omega$. Then, does there exist $x\in G$ such that $x\ast x=e$ and $x\neq e$?

MJD
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John. p
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3 Answers3

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Depending on what you know, the answer can be almost painfully trivial: yes, as any group of even order has an element of order two by Cauchy's Theorem.

If you haven't yet covered the above theorem things are going to be lengthier: pair up as many as possible of the $\;2n\;$ elements of the group by the rule $\;(x\,,\,x^{-1})\;$ . Taking into account that the unit element gets paired with itself, and since the number of elements of the group is even, there must be another, non-unit!, element in the group which is paired with itself, and we're done.

DonAntonio
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To expand on Daniel Fischer's hint:

Let $f : G \to G$ be defined by $f(x) = x^{-1}$. This is a bijection. An element $x \in G$ verifies $x * x = e$ iff $f(x) = x$, so you want to find the fixed points of $f$, or at least find one that is not $e$. Note that $f(e) = e$ already, so that's one fixed point.

Now $f$ is an involution ($f(f(x)) = x$), and $|G|$ is even, so its number of fixed points has to be even too: let $F \subset G$ be the fixed points, then every element $x \in G \setminus F$ can be paired with $f(x)$, thus $|G \setminus F|$ is even, so $|F|$ is even too. And since we already have one fixed point ($e$) then there is at least one other (because $1$ is odd), say $x \neq e$. Then $x$ has order $2$ and we're done.

Najib Idrissi
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  • the very question points towards the plausibility that the OP's level in group theory is pretty elementary. What are the odds he has studied groups actions on sets, fixed points and etc.? – DonAntonio Mar 25 '14 at 13:48
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    @DonAntonio There's nothing with group actions here. Just the very elementary fact that the number of fixed points of an involution has the same parity as the cardinality of the set. – Daniel Fischer Mar 25 '14 at 13:51
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    Even so, @DanielFischer (though with group actions it would also be pretty straightforward): involutions, fixed points and etc. are notions that are not usually covered, as far as I can tell, at the beginning of group theory courses. They belong more towards subjects like automorphisms groups, actions and stuff. And the "very elementary" fact you mention may not seem elementary at all to a beginner in these matters. – DonAntonio Mar 25 '14 at 13:53
  • @DonAntonio Fixed points are a very basic notion independent of algebra, if somebody hasn't encountered them before taking a course on algebra, I would regard that as a very weird curriculum. Involutions are also not quite unimportant outside algebra. I would however be only mildly surprised if one didn't know what an involution is when taking the first course on algebra. However, that $\iota\colon x \mapsto x^{-1}$ is its own inverse is obvious. – Daniel Fischer Mar 25 '14 at 13:57
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    @DanielFischer the only other basic stuff in the general mathematics curriculum that includes fixed points I can think of is calculus...and not very basic one: Brouwer's or Banach's fixed point theorems aren't basic, imo. Other things could include numerical approximations (Newton's and etc.), but I'd say "fixed points of involutions" is a neat, almost pure notion belonging to abstract algebra (or even a little to linear algebra with linear maps and eigenvalues equal to one...), and in my expererience none is very-very elementary at all. – DonAntonio Mar 25 '14 at 14:01
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    @DonAntonio Brouwer's isn't basic (except in dimension $1$), but Banach's is pretty basic. Nothing but metric spaces and completeness involved, used to prove the inverse and implicit function theorems, second semester stuff. Algebra comes fourth semester hereabouts. – Daniel Fischer Mar 25 '14 at 14:09
  • Banach's basic...in Calculus III or Functional analysis I, @DanielFischer. The stuff you mention isn't usually covered, at least where I know a little (Israel, Mexico and USA) in the first 1-2 years. If at all, towards the end of a third or four semester in Calculus and Advanced Calculus. You can check this in at least some of the basic calculus texts out there (though perhaps not all). We here begin group theory in third semester, and in 4th we do fields extensions and Galois Theory, plus some optional courses in commutative algebra and etc. – DonAntonio Mar 25 '14 at 14:58
  • @DanielFischer, and the Implicit Function theorem...in second semester? How fast do you first cover limits of sequences, functions, series, derivative, integration, Taylor and power series and beginning of multivariable calculus?? – DonAntonio Mar 25 '14 at 14:59
  • @DonAntonio First semester (Analysis I) is limits of sequences and series, continuity, differentiation (including Taylor series) and Riemann integral (all in dimension $1$) plus a little about Fourier series. Second semester is multivariable analysis, partial differentiation, total differential, a wee bit of topology (metric spaces, though almost exclusively Euclidean spaces and normed spaces over the real or complex numbers), inverse and implicit function theorems, Lagrange multipliers etc. If there's enough time, one starts ODEs in the second semester, otherwise in the third. – Daniel Fischer Mar 25 '14 at 15:06
  • Third semester is ODEs and integration in dimensions $> 1$, Lebesgue integral, plus a little analysis on submanifolds of $\mathbb{R}^n$. – Daniel Fischer Mar 25 '14 at 15:07
  • @Daniel I agree that Banach's fixed point theorem is quite basic, but really are Banach fpt&inverse&implicit function theorem covered in second semetester? I saw three~four elementary calculus texts and they all merely stated those theorems without proof. – John. p Mar 25 '14 at 15:24
  • @John.p In Germany, they were at least. Things have recently been reorganised when Bachelor and Master were introduced, I don't know how that affected the curriculum. Here, the rule was that a result could not be used before it was proved (in the first semesters; later, it was assumed the students are grown up enough to be able to deal with "we need that result but haven't the time to prove it, read it up in the literature"). I gather that there are places where the studies are organised in a different way. I still have problems with the idea of teaching mathematics without proofs. – Daniel Fischer Mar 25 '14 at 15:35
  • Same rule exists here: whatever you use that wasn't covered in class you must prove it. I thought we were fast here in Israel as the B.Sc. title is only three years (!) and then M.Sc. and Ph.D., but we definitely do not cover that much in analysis in first and second semester...not even close! We though covered quite a bit of topology and even cardinality and stuff in first semester, but Fourier series is definitely 3rd semester and +. And also I would never be able to teach mathematics without proofs, but: is there any place in this sector of the galaxy where they do such a thing?? – DonAntonio Mar 25 '14 at 16:38
  • My apologies, but would you mind moving this discussion somewhere else? Chat for example? I keep getting notifications :/ – Najib Idrissi Mar 25 '14 at 19:17
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(This is a variation on Daniel Fischer's hint in the comments.)

Let's partition the elements of $G$ into subsets: For each $x\in G$, there will be a subset $S_x$ consisting of exactly $\{x, x^{-1}\}$. Note that when $x =x^{-1}$, this subset will have one element, not 2. For example, $S_e$, the subset containing $e$ is $\{e\}$.

Now prove the following:

  1. This division into subsets is a partition: each element of $G$ is in exactly one of the subsets.

  2. If you add up the sizes of the subsets, you get $2n$.

  3. Some of the subsets have 2 elements, some have 1. But there is at least one subset with only 1 element.

By (3), there is at least one subset with only one element. Is it possible that there are no others?

MJD
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  • Why is it impossible that every partition is a singletone? – John. p Mar 25 '14 at 13:48
  • It's not impossible, @John.p. But then you're also done, all you need is at least one singleton in the partition besides ${e}$. – Daniel Fischer Mar 25 '14 at 13:52
  • @John.p It is quite possible that they are all singletons. My question is “Is it possible that there is only one singleton?” – MJD Mar 25 '14 at 13:55