4

The exercise asks to prove that if $G$ is any group with ${\rm Aut}(G) = \{{\rm id}\}$, then $g^2=1$ for all $g$ in $G$, $G$ is abelian, and if $G$ is finite, we'll have $|G| = 1$ or $2$.

I managed to do everything up to the last part (and so far, we can find good answers on this site), but I don't know how to prove that if $|G|>1$, then $|G|=2$. I just know that $|G|$ is even. Can someone give hints? Thanks.

Ivo Terek
  • 77,665
  • If you can prove that last statement, that $|G|\le 2$, then the first two statements follow right away. $\qquad$ – Michael Hardy Mar 18 '16 at 18:19
  • 3
    Hint: conjugation by any element is an automorphism. The statement $gxg^-1 = x$ for all x and g in G is equivalent to G being abelian. What special automorphisms do abelian groups have that nonabelian groups do not? – Vik78 Mar 18 '16 at 18:39
  • Well, the only automorphism that comes to mind is $x \mapsto x^{-1}$, but this gives us $x^2 = 1$, which we already knew.. and $x\mapsto x^2$ will be a morphism but maybe not an automorphism. I read the answer linked above, but we haven't covered stuff about $GF(2)$ and such, so maybe there is an easy way to solve that – Ivo Terek Mar 18 '16 at 19:19
  • 1
    Have you done the finite abelian group structure theorem yet? If G is finite then the fact that every non identity element has order 2 implies that it's a direct product of several copies of the group of order 2 (or the trivial group). But if it was a product of more than one copy then interchanging the generators of any two copies would be a nontrivial automorphism. – Vik78 Mar 18 '16 at 19:26
  • @Vik Sorry if I'm slow. Doesn't the theorem says that it will be a direct product of groups of order a power of $2$ too? Since every non-identity element has order $2$, then $|G| = 2^n$ for some $n$ and then $$G \cong \Bbb Z_{2^{n_1}}\oplus \cdots \oplus \Bbb Z_{2^{n_t}}$$where $\sum n_j = n$. I understood that automorphism, but it will imply that $G \cong \Bbb Z_{2^n}$. But so what? – Ivo Terek Mar 18 '16 at 19:38
  • But if there is a group of a power of 2 greater than 2 itself in G's factor decomposition, then the independent generator of that factor has order greater than 2. – Vik78 Mar 18 '16 at 19:47
  • @Vik Ah, of course. Ok. Then we get $G \cong \Bbb Z_2$ and everybody is happy. Thanks for the patience. If you want to join your comments in an answer, I'll accept it. – Ivo Terek Mar 18 '16 at 19:57

2 Answers2

3

Given $g\in G$, Since conjugation by $g$ is an automorphism of $G$, we have that for all $h\in G$, $ghg^{-1}=h$ and so $gh=hg$. This shows that $G$ is abelian. Now, since $G$ is abelian, the map $x\mapsto x^{-1}$ is an automorphism of $G$, and by hypothesis, $g^{-1}=g\Rightarrow g^2=1$ for all $g\in G$.

Finally, let $S=\{a,b,c_1,\ldots,c_k\}$ be a set of generators of $G$ with $a\neq b$. Notice that the obtained results implies that every element $x\in G$ can be expressed as a product $x=a^ib^jc_x$ where $i,j\in \{0,1\}$ and $c_x$ is a word only using $c_1,\ldots,c_k$ (each one at most once). Let us write furthermore $x'=a^ib^j$.

Consider the permutation $\sigma:=\begin{pmatrix}e&a&b&ab\\e&b&a&ab\end{pmatrix}$. Note that $\sigma$ is an automorphism of the subgroup $G'=\langle a,b\rangle$. Define now the function $\phi:G\to G$ given by

$$\phi(x)=\phi(x'c_x):=\sigma(x')c_x.$$

The map $\phi$ is a bijection, and given $x,y\in G$ we have:

$$\phi(x\cdot y)=\phi(x'y'c_xc_y)=\sigma(x'y')c_xc_y=\sigma(x')\sigma(y')c_xc_y=\phi(x)\cdot\phi(y)$$

So, $\phi$ is an automorphism of $G$, and by hypothesis, we must have $a=\phi(a)=\sigma(a)=b$. Contradiction.

Darío G
  • 4,878
2

If you've managed to do everything up to the last part, then you know that $G$ is an abelian group for which $g^2=e$ for all $g\in G$. This means that $G$ is an $\mathbb{F}_2$-vector space, so you have lots of automorphisms to work with if $|G|\neq 1,2$.

carmichael561
  • 53,688
  • Additional question: can you give an example of an infinite group with only one automorphism? – Vik78 Mar 18 '16 at 20:22
  • @Vik78 There is none by the same argument. But in the infinite case, you need to use the fact that every (infinite-dimensional) $\mathbb{F}_2$-vector space has a basis, which is probably beyond the scope of the course in which this exercise was assigned. This fact is a consequence of the axioms of choice, but if we do not assume choice there may be such a group. – Alex Kruckman Mar 18 '16 at 20:49