This question already has an answer here:

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 $x\neq e$?


marked as duplicate by Jyrki Lahtonen Jun 20 '15 at 8:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ How many fixed points does the map $x\mapsto x^{-1}$ have? $\endgroup$ – Daniel Fischer Mar 25 '14 at 13:32
  • $\begingroup$ @Daniel Is that a hint or are you really asking? The question in the post is exactly an exercise in my textbook.. $\endgroup$ – John. p Mar 25 '14 at 13:35
  • $\begingroup$ A hint. You don't need the exact number, by the way. $\endgroup$ – Daniel Fischer Mar 25 '14 at 13:36
  • $\begingroup$ @luli Is that a hint? I don't get it.. Would you give me some more details? $\endgroup$ – John. p Mar 25 '14 at 13:36
  • $\begingroup$ @Daniel Would you give me more details? I understand that map is an automorphism, but then next i dunno what to do $\endgroup$ – John. p Mar 25 '14 at 13:38

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.

  • $\begingroup$ 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.? $\endgroup$ – DonAntonio Mar 25 '14 at 13:48
  • 1
    $\begingroup$ @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. $\endgroup$ – Daniel Fischer Mar 25 '14 at 13:51
  • 1
    $\begingroup$ 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. $\endgroup$ – DonAntonio Mar 25 '14 at 13:53
  • 1
    $\begingroup$ @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. $\endgroup$ – DonAntonio Mar 25 '14 at 14:01
  • 1
    $\begingroup$ @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. $\endgroup$ – Daniel Fischer Mar 25 '14 at 14:09

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.


(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?

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.