Let $G$ be a finite group with $|G|>2$. Prove that ${\rm Aut}(G)$ contains at least two elements.

Question

Let $G$ be a finite group with $|G|>2$. Prove that ${\rm Aut}(G)$ contains at least two elements.

We know that ${\rm Aut}(G)$ contains the identity function $f: G \to G: x \mapsto x$.

If $G$ is non-abelian, look at $g : G \to G: x \mapsto gxg^{-1}$, for $g\neq e$. This is an inner automorphism unequal to the identity function, so we have at least two elements in ${\rm Aut}(G).$

Now assume $G$ is abelian. Then the only inner automorphism is the identity function. Now look at the mapping $\varphi: G \to G : x \mapsto x^{-1}$. This is an homomorphism because $\varphi (xy) = (xy)^{-1} = y^{-1} x^{-1} = x^{-1} y^{-1} = \varphi (x) \varphi (y)$. Here we use the fact that $G$ is abelian. This mapping is clearly bijective, and thus an automorphism.

This automorphism is unequal to the identity function only if there exists an element $x \in G$ such that $x \neq x^{-1}$. In other words, there must be an element of order greater than $2$.

Now assume $G$ is abelian and every non-identity element has order $2$. By Cauchy's theorem we know that the group must have order $2^n$.

I got stuck at this point. I've looked at this other post, $|G|>2$ implies $G$ has non trivial automorphism, but I don't know what they do in the last part (when they start talking about vector spaces). How should this prove be finished, without resorting to vector spaces if possible?

Thanks in advance

Answer 1

The reason why that last case is done separately is because in that case we cannot describe a non-trivial automorphism with operations intrinsic to the group. In the non-abelian case we could use an inner automorphism, and in other abelian cases we can use negation. Using vector space structure is the "next best thing" in the sense that then everybody will immediately accept multiplication by an invertible matrix as an automorphism.

That last case is about groups that are isomorphic to a direct product of finitely many copies of $C_2$ $$ G=C_2\times C_2\times\cdots C_2. $$ Any permutations of components is then obviously an automorphism. But proving that such a group has this direct product structure, while not too tricky, is something the answerers did not want to do, because it is much more convenient to observe that a finite abelian group where all the non-trivial elements have order two is a vector space over $GF(2)$.

Here's another way of doing this step. We inductively find elements $x_1,x_2,\ldots,x_\ell\in G$ such that at all steps $x_k$ is not in the subgroup generated by $x_1,x_2,\ldots,x_{k-1}$. Because $G$ is finite, at some point we will have exhausted all of $G$, and thus $G=\langle x_1,x_2,\ldots,x_\ell\rangle$. The claim is then that $$ G=\langle x_1\rangle\times\langle x_2\rangle\times\cdots\langle x_\ell\rangle $$ is a way of writing $G$ as a direct product of subgroups of order two. This is more or less obvious by construction.

But you should notice that this is essentially the same argument proving that a finitely generated vector space over a field has a finite basis.

Answer 2

By structure theorem for finite abelian groups, we have $$G\cong(\Bbb Z_2)^n$$ for some $n$. Consider the function on $(\Bbb Z_2)^n$ that swaps the $1$st and $n$th coordinates. Show that that is an automorphism.

Answer 3

Building off of OP's work, we now consider Abelian groups, where each non-identity element has order 2.

Let $\{g_1, g_2, \ldots g_n \}$ be any minimal generating set of the group. Since $|G| > 2 $, thus $n\geq 2$.

Each element of the group $g$ can be uniquely written as $\prod_{i \in S_g} g_i$. (Exponents are no needed, since each element has order 2.)

Then, consider the the map that replaces $g_1$ with $g_2$ and vice versa.

Show that this is an automorphism.