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I am reading "An Introduction to Algebraic Systems" by Kazuo Matsuzaka.

There is the following problem in this book:

Let $G$ be a non-abelian finite group.
Let $f$ be an automorphism of $G$ such that $f^2 = I_G$.
Show that there exists $x \in G$ such that $f(x) = x$ and $x \neq e$.

My attempt is here:

(1)
If $x' = f(x)$, then $x = f(f(x)) = f(x')$.
So, we can write as follows:
$x \stackrel{\mathrm{f}}{\longleftrightarrow} x'$.
This problem says there exists $y \neq e$ such that
$y \stackrel{\mathrm{f}}{\longleftrightarrow} y$.

(2)
$H := \{x \in G | f(x) = x\}$ is a subgroup of $G$.
So, we need to show $H \neq \{e\}$.

(3)
Let $G = \{x_1, x_2, \cdots, x_n\}$.
Let $f(x_i) = x_i'$ for $i \in \{1, 2, \cdots, n\}$.

Then,

$x_1 \stackrel{\mathrm{f}}{\longleftrightarrow} x_1'$.
$x_2 \stackrel{\mathrm{f}}{\longleftrightarrow} x_2'$.
$\cdots$
$x_n \stackrel{\mathrm{f}}{\longleftrightarrow} x_n'$.

And

$G = \{x_1, x_2, \cdots, x_n\} = \{x_1', x_2', \cdots, x_n'\}$.

So,

$x_1 x_2 \cdots x_n \stackrel{\mathrm{f}}{\longleftrightarrow} x_1' x_2' \cdots x_n' = x_1 x_2 \cdots x_n$.

So,

$x_1 x_2 \cdots x_n \in H$.

But I cannot show that $x_1 x_2 \cdots x_n \neq e$.

(4)
I don't use the assumption that $G$ is non-abelian yet.

tchappy ha
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3 Answers3

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Suppose $x_1 x_2 \neq x_2 x_1$. Then $x_2x_1...x_n\stackrel{\mathrm{f}}{\longleftrightarrow} x'_2x'_1...x'_n=x_2x_1...x_n$. Since $x_1x_2...x_n \neq x_2x_1...x_n$, we must have $H \neq \{e\}$.

Lázaro Albuquerque
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Use the line of reasoning in Show that a finite group with certain automorphism is abelian by Arturo Magidin [AM] and user9413 [U]: If the only fixed point of $f$ is $e$ then $G$ is abelian.

However, I found the first step to show that showing Claim 1 below [AM] [U], to be nontrivial so I worked out the proof for myself.

Claim 1: If $f$ as specified has only $\{e\}$ as the fixed point, then every $g \in G$ can be written as $x^{-1}f(x)$ for some $x \in G$.

So here is my proof of Claim 1: Suppose $x$ and $y$ in $G$ satisfy $x^{-1}f(x) = y^{-1}f(y)$ [AM]. Then applying $f$ to both sides yields:

$f(x^{-1}f(x)) = f(x^{-1})f(f(x)) = f(x^{-1})x = f(y^{-1})y$.

Then this implies $f(y)f(x^{-1}) = yx^{-1}$, which implies $f(yx^{-1}) = yx^{-1}$, which would imply that $yx^{-1}$ is a fixed point of $f$, which, by the assumption that $e$ is the only fixed point, implies $yx^{-1} = e$, which implies $y=x$. Thus $h: x \mapsto x^{-1}f(x)$ is a one-to-one of $G$ onto itself, and [as $G$ is finite], implies that the image of $h$ is $G$ itself, implying Claim 1.

Mike
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  • Why are you phrasing that argument as a proof by contradiction? It's a direct proof! "If $x^{-1}f(x)= y^{-1}f(y)$, then $x=y$." All you've done is add a "assume that $x\neq y$" at the top, and add "but that contradicts our assumption that $x\neq y$" at the bottom of the proof, to make it longer and less clear. – Arturo Magidin Jan 28 '19 at 16:29
  • Why would that be? I am not seeing that – Mike Jan 28 '19 at 16:30
  • The way I did it out was the way I saw the claim. It was otherwise not obvious to me that it was true – Mike Jan 28 '19 at 16:32
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    What you have is a "fake proof by contradiction". It's actually a direct proof, dressed up to include a superfluous hypothesis at the top. Here's your argument, as a direct proof: "Suppose that $x,y\in G$ are such that $x^{-1}f(x)=y^{-1}f(y)$. Applying $f$ to both sides yields [...] $f(yx^{-1})=yx^{-1}$. Hence, $yx^{-1}=e$, so $y=x$. Therefore, $f$ is one to one." And since $G$ is finite, $f$ being one-to-one implies it is onto, which gives you the statement. You don't have to assume $x\neq y$, which is never used in your argument, until you say "but that's a contradiction." – Arturo Magidin Jan 28 '19 at 16:33
  • We cross-posted. I justed editted my proof will check now to make sure it is the same as yours above – Mike Jan 28 '19 at 16:35
  • Well, you stated your claim as an assertion that the map is onto; you proved that it is one-to-one. You want to connect the two. Otherwise, yes, you've removed the "fake-proof-by-contradiction" part. As to that term, see the paragraph after "That said" here: https://math.stackexchange.com/questions/112774/proof-by-contradiction-circular-reasoning – Arturo Magidin Jan 28 '19 at 16:37
  • OK I see it now. I did this out admittedly for my sake too. – Mike Jan 28 '19 at 16:39
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Thank you very much, Izralbu, Tsemo Aristide, Mike.

I write this answer for my memo.

If $x' = f(x)$, then $x = f(f(x)) = f(x')$.
So, we can write as follows:

$x \stackrel{\mathrm{f}}{\longleftrightarrow} x'$.

This problem says there exists $y \neq e$ such that

$y \stackrel{\mathrm{f}}{\longleftrightarrow} y$.

(1)
Assume that there doesn't exist $y \neq e$ such that

$y \stackrel{\mathrm{f}}{\longleftrightarrow} y$.

(2-1)
We show that $G \ni x \longmapsto x^{-1} x' \in G$ is injective.

Assume that $x_1^{-1} x_1' = x_2^{-1} x_2'$.
Then, $x_1 x_2^{-1} = x_1' x_2'^{-1}$.

(2-2)
$x_1 \stackrel{\mathrm{f}}{\longleftrightarrow} x_1'$.
$x_2^{-1} \stackrel{\mathrm{f}}{\longleftrightarrow} x_2'^{-1}$.

So, $x_1 x_2^{-1} \stackrel{\mathrm{f}}{\longleftrightarrow} x_1' x_2'^{-1} = x_1 x_2^{-1}$ by (2-1).

(2-3)
By assumption (1), $x_1 x_2^{-1} = e$.
So, $x_1 = x_2$.
So, $G \ni x \longmapsto x^{-1} x' \in G$ is injective.

(3)
By assumption, $G$ is finite.
So, $G \ni x \longmapsto x^{-1} x' \in G$ is surjective.
So, $G \ni x \longmapsto x^{-1} x' \in G$ is bijective.

(4)
$\{x_1, x_2, \cdots, x_n\} = \{x_1^{-1} x_1', x_2^{-1} x_2', \cdots, x_n^{-1} x_n'\}$ by (3).

(5)
Let $g$ be an arbitrary element of $G$.
Then, there exists $h \in G$ such that $g = h^{-1} h'$.
Then, $g = h^{-1} h' \stackrel{\mathrm{f}}{\longleftrightarrow} h'^{-1} h = (h^{-1} h')^{-1} = g^{-1}$.

(6)
Let $x, y$ be arbitrary elements of $G$.
$x \stackrel{\mathrm{f}}{\longleftrightarrow} x^{-1}$ by (5).
$y \stackrel{\mathrm{f}}{\longleftrightarrow} y^{-1}$ by (5).
So, $x y \stackrel{\mathrm{f}}{\longleftrightarrow} x^{-1} y^{-1}$.
And, $x y \stackrel{\mathrm{f}}{\longleftrightarrow} (x y)^{-1}$ by (5).
$\therefore x^{-1} y^{-1} = (x y)^{-1}$.
$\therefore x^{-1} y^{-1} = y^{-1} x^{-1}$.
$\therefore x y = y x$.
$\therefore G$ is abelian.

This contradicts the assumption that $G$ is non-abelian.

tchappy ha
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