MHB Can a Group Have a Trivial Automorphism Group with Less than Three Elements?

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Assuming the axiom of choice, show that a group $G$ has trivial automorphism group if and only if $G$ has less than three elements.

 

[mathbalarka]

Spoiler

After considerable help from http://math.stackexchange.com/users/98602/mike-miller of MSE (with whom I use to have frequent conversations), I finally solved it.

One impliction is patently obvious, as if $|G| < 3$, then $|G|$ is either $1$, in which case it is straightforward that automorphism of trivial groups is trivial, or $2$ in which case $G \cong \Bbb Z/2\Bbb Z$ and the only automorphism of this group is $x \mapsto x$. For the other implication, we divide the proof into two cases, in which $G$ is abelian and where $G$ is nonabelian :

Claim : $Aut(G)$ is nontrivial for $|G| \geq 3$, given $G$ is abelian.

Proof : As $|G|$ is at least $3$, it contains $1$, $x$ and the inverse $x^{-1}$ of $x$. The map $x \mapsto x^{-1}$ is an automorphism and it's trivial if and only if $x^2 = 1$ for all $x \in G$. Assume it is not the case, then $x \mapsto x^{-1}$ is the desired nontrivial auto of $G$.

If, on the other hand, every element of $G$ has order $2$, then $G$ is a vector space over $\Bbb Z/2\Bbb Z$, i.e., $G = \bigoplus \Bbb Z/2\Bbb Z$. Consider the map $(x_1, x_2, x_3, \cdots) \mapsto (x_2, x_1, x_3, \cdots)$. This is always possible for finite $G$, as there are at least two copies of $\Bbb Z/2\Bbb Z$ sitting inside for all $|G| > 2$. However, if $G$ is infinite, picking two elements of $\Bbb Z/2\Bbb Z$ from infinitely many of the elements requires Zorn's lemma (i.e., Axiom of Choice in disguise). So we have proved that if $G$ is abelian, the statement of the problem holds, conditional on AC.

Claim : $Aut(G)$ is nontrivial for $|G| \geq 3$, given $G$ is nonabelian.

Proof : Consider the conjugation map $x \mapsto axa^{-1}$. This is clearly an automorphism, and it is trivial if and only if $axa^{-1} = x$ which happens if $a$ commutes with $x$. But as $Z(G)$ is nontrivial by assumption, there exists a choice of $a$ such that $ax \neq xa$ for at least one $x \in G$, which proves the existence of a nontrivial map $x \mapsto axa^{-1}$.

This concludes the proof $\blacksquare$

 

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Hi mathbalarka,

Spoiler

There is incoherence in your proof that needs revision.

First, assume that $G$ is abelian. As $|G|$ is at least $3$, it contains $1$, $x$ and the inverse $x^{-1}$ of $x$.The map $x \mapsto x^{-1}$ is an automorphism and it's trivial if and only if $x^2 = 1$ for all $x \in G$. Assume it is not the case, then $x \mapsto x^{-1}$ is the desired nontrivial auto of $G$.

It does not follow from the abelian property of $G$ that $|G| \ge 3$. If you're giving a proof by contradiction, then make that clear from the beginning. Since $x$ is a particular point of $G$, the assignment $x \mapsto x^{-1}$ is not defined as a function from $G$ into $G$.

If, on the other hand, every element of $G$ has order $2$, then $G$ is a vector space over $\Bbb Z/2\Bbb Z$, i.e., $G = \bigoplus \Bbb Z/2\Bbb Z$. Consider the map swapping two of the $\Bbb Z/2\Bbb Z$s sitting inside, which is an automorphism. This is always possible for finite $G$, as there are at least two copies of $\Bbb Z/2\Bbb Z$ sitting inside for all $|G| > 2$. However, if $G$ is infinite, picking two copies of $\Bbb Z/2\Bbb Z$ from infinitely many of the copies requires Zorn's lemma (i.e., Axiom of Choice in disguise). So we have proved that if $G$ is abelian, the statement of the problem holds, conditional on AC.

You have good ideas, but this part is vague and not carefully written. You can't swap copies in $G$, but elements in $G$.

The first and second paragraphs are not unified. Since the original statement is biconditional you also need to prove the other direction.

 

[mathbalarka]

Spoiler

It does not follow from the abelian property of G that |G|≥3.

You didn't quite get my point. What I have shown is that $Aut(G)$ is nontrivial for $|G| \geq 3$ assuming $G$ is abelian.

Since x is a particular point of G, the assignment x↦x−1 is not defined as a function from G into G.

Huh? The map $G \to G$ as $x \mapsto x^{-1}$ is always defined on $G$. Every element of a group has a unique inverse.

You can't swap copies in G, but elements in G.

I am aware of what I am doing. I believe it is standard terminology to say that $(a, b) \mapsto (b, a)$ swaps the copies of $A$ and $B$.

Seriously, if that's proof writing and very formal terminologies you need instead of the mathematics, I am not going to try writing them out.

EDIT I have edited my post above to make stuffs more clear.

 

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Your proof looks much better now. Thanks for participating, mathbalarka!

 

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Spoiler

Suppose $G$ has less than three elements. Then either $G$ is trivial or isomorphic to $\Bbb Z/2\Bbb Z$. In both cases, $\text{Aut}(G)$ is trivial.

The converse will be proven by contradiction. Suppose $G$ has at least three elements. Since $\text{Inn}(G) \le \text{Aut}(G)$ and $\text{Aut}(G)$ is trivial, so is $\text{Inn}(G)$. Therefore, $G$ is abelian. The inversion map is therefore an element of $\text{Aut}(G)$, whence $G$ has exponent two. So $G$ is an $\Bbb F_2$-vector space. Let $e$ be the identity element of $G$. Take distinct elements $g, h \in G\setminus\{e\}$. Using the axiom of choice, choose a basis $\mathcal{B}$ for $G$ that contains $g$ and $h$. Let $T : \mathcal{B} \to \mathcal{B}$ be function such that $T(g) = h$, $T(h) = g$, and $T(x) = x$ for all $x \in \mathcal{B} \setminus\{g,h\}$. Then $T$ extends to a linear isomorphism $\tilde{T}: G \to G$, which is a nontrivial element of $\text{Aut}(G)$. This is a contradiction.