27

If $G$ is a group of even order, prove it has an element $a \neq e$ satisfying $a^2 = e$.

My proof:

Let $|G| = 2n$. Since $G$ is finite, there exists, $a \in G$ such that $a^p = e$ and by Lagrange's Theorem, p divides 2n. By Euclid's lemma, since p does not divide 2, p divides n. Let $n = pk$. Hence, $(a^n)^2 = (a^{pk})^2 = ((a^p)^k)^2 = (e^k)^2 = e$. Therefore, $a^n$ is an element that satisfy the condition.

Is my solution OK?

For this problem, I am just wondering how I can solve this problem without using Lagrange's Theorem, as this problem is an exercise before the Lagrange's Theorem was taught.

Shaun
  • 44,997
user10024395
  • 1
  • 1
    I don't see any problem. – kurtzdoni Jan 20 '15 at 11:53
  • 9
    You must exclude that $a^n = e$. – AlexR Jan 20 '15 at 11:56
  • Check the list of Related questions (on the right), and you'll find some insights. – Blue Jan 20 '15 at 11:57
  • @AlexR what do you mean by i must exclude $a^n = e$ – user10024395 Jan 20 '15 at 12:01
  • @user136266 If $a^n = e$, you only show $e^2 = e$ in a way. – AlexR Jan 20 '15 at 12:03
  • @user136266 Here is a correct proof if the statement. – AlexR Jan 20 '15 at 12:04
  • @AlexR if $a^n = e$, it doesn't mean that $a = e$, does it? – user10024395 Jan 20 '15 at 12:12
  • 1
    @user136266 But you say "$a^n$ is an element that satisfies the condition", implying $a^n \ne e$ without first ruling out $a^n = e$. – AlexR Jan 20 '15 at 12:14
  • @AlexR I think you (should) mean that the case $;a=e;$ must be ruled out, otherwise there is no proof. Also, $;a=e=e^1=e;$ and indeed $;1\mid 2;$, and the OP's proof falls apart. – Timbuc Jan 20 '15 at 13:41
  • @Timbuc $a^n = e$ already destroys the proof, wich is especially the case for $a=e$. – AlexR Jan 20 '15 at 13:43
  • 1
    This whole argument is a result of user13626 using $a$ in two roles: the $a$ of the first sentence ("prove it has an element $a \ne e...$") is different from the $a$ in the proof ("there exists, $a \in G...$"). You should say "Take any $x \in G$ such that $x^n \ne e$. Then there exists $p \in \mathbb N$ such that..." Then you end up choosing $a=x^n$. You also have to show that there exists such an $x$ (with $x^n \ne e$). – TonyK Jan 20 '15 at 13:58
  • 1
    $a^n$ has order 1, because it is always $e$. – IS4 Mar 18 '19 at 13:35
  • I am a little puzzled by what $p$ is supposed to be. Here is something along these lines. Start by taking a non-trivial element $g\in G$ which has order dividing $2n$. If its order $d$ is even then $g^{d/2}$ has order 2. Otherwise its order is odd so $d=pd'$ where $p$ is an odd prime. But now I don't see how to go further. I think the proofs given by other people are the way to go. Of course the general version for a prime dividing $|G|$ is Cauchy's Theorem and there are many proofs. – Andy Baker Jan 09 '22 at 13:29

2 Answers2

85

The following is perhaps one of most simple proofs:

Pair up if possible each element of $\;G\;$ with its inverse, and observe that

$$g^2\neq e\iff g\neq g^{-1}\iff \;\text{there exists the pair}\;\;(g, g^{-1})$$

Now, there is one element that has no pairing: the unit $\;e\;$ (since indeed $\;e=e^{-1}\iff e^2=e$), so since the number of elements of $\;G\;$ is even there must be at least one element more, say $\;e\neq a\in G\;$ , without a pairing, and thus $\;a=a^{-1}\iff a^2=e\;$

Timbuc
  • 34,191
  • 1
    @user136266 Who said anything about "only" two such elements? The above proves that besides the unit $;e;$ there must be another element with $;a^2=e;$, since the number of elements in $;G;$ is even. In fact, the above shows that under the conditions of the problem, there is always an odd number of non-unit elements which have order two. – Timbuc Jan 20 '15 at 14:46
  • 1
    Very neat! ${}{}$ –  May 05 '17 at 18:13
13

This is not a strict proof, but you may find it helpful when you want proof without Lagrange's theorem:

We have that for every $g\in G$ there a unique $g^{-1} \in G$ such that $gg^{-1}=e$. If you suppose that there is no $a \in G$ such that $a^2=e$, so that $a=a^{-1}$ (i.e. there is no self-inverse element), then for every $x\neq e$ in $G$ we can assign unique $y\in G$ such that $xy=e$. So the set of pairs of elements that are inverses to each other form a partition for $G$.

But then $|G|$ is odd since $e$ is only self-inverse element.

user160738
  • 4,160
  • You need to exclude the case $p = n$ also (in the second argument). – Krish Jan 20 '15 at 12:32
  • If $p$ is the order of $a$ then $a^n = a^{pk} = e^k = e$. $a^n$ will never be the required element. – AlexR Jan 20 '15 at 13:46
  • Smallest counter-example is any element of order $3$ in $(\mathbb Z_{12}, +)$. – AlexR Jan 20 '15 at 14:05
  • 1
    "Then either $p$ divides $2$ or it doesn't since $2$ is a prime": In my universe, "either $p$ divides $n$ or it doesn't" is true for composite $n$ too :-) – TonyK Jan 20 '15 at 14:06
  • @AlexR yep I wasn't aware of that when I was writing. I cannot think of a way to refine this style of proof (proving existence by finding one) to make it true unfortunately :( – user160738 Jan 20 '15 at 14:47
  • @TonyK Haha true I must have said that "either $p$ divides $2$ or $n$ since $2$ is a prime – user160738 Jan 20 '15 at 14:50
  • @user160738 Since there is already an accepted answer and the original proof is unsalvagable, I suggest you delete the bottom part your answer and undelete it once you found a construction wich actually works. – AlexR Jan 20 '15 at 14:53
  • @AlexR Done it! – user160738 Jan 20 '15 at 14:58