Let \(G\) be a finite group and let \(\varphi\in\operatorname{Aut}(G)\) fix only the identity element.
Show that every element of \(G\) has the form \[ g^{-1}\varphi(g) \] for some \(g\in G\).
If \(\varphi\) has order \(2\), show that \(\varphi(g)=g^{-1}\) for every \(g\in G\), and deduce that \(G\) is abelian of odd order.
Solution
Define \[ \psi:G\longrightarrow G, \qquad \psi(g)=g^{-1}\varphi(g). \]
1. The map \(\psi\) is injective.
Suppose \[ \psi(g)=\psi(h). \] Then \[ g^{-1}\varphi(g) = h^{-1}\varphi(h). \] Multiplying on the left by \(h\) and on the right by \(\varphi(g)^{-1}\) gives \[ hg^{-1} = \varphi(h)\varphi(g)^{-1}. \] Because \(\varphi\) is a homomorphism, \[ \varphi(h)\varphi(g)^{-1} = \varphi(hg^{-1}). \] Thus \(hg^{-1}\) is fixed by \(\varphi\). The only fixed point is the identity, so \[ hg^{-1}=e, \] and hence \(h=g\).
2. The map \(\psi\) is surjective.
The group \(G\) is finite. An injective self-map of a finite set is surjective, so step 1 gives surjectivity.
3. Every element of \(G\) can be written in the form\[ \boxed{ g^{-1}\varphi(g) } \]for some \(g\in G\).
This is exactly surjectivity of \(\psi\) from step 2.
4. Assume now that\[ \varphi^2=\operatorname{id}. \]Then for every \(x\in G\),\[ \boxed{ \varphi(x)=x^{-1}. } \]
By step 3, write \[ x=g^{-1}\varphi(g) \] for some \(g\in G\). Then \[ \begin{aligned} \varphi(x) &= \varphi(g)^{-1}\varphi^2(g)\\ &= \varphi(g)^{-1}g. \end{aligned} \] But \[ x^{-1} = \bigl(g^{-1}\varphi(g)\bigr)^{-1} = \varphi(g)^{-1}g. \] Hence \(\varphi(x)=x^{-1}\).
5. The group \(G\) is abelian.
By step 4, the inversion map is the automorphism \(\varphi\), so it is a homomorphism. Therefore for all \(g,h\in G\), \[ (gh)^{-1} = g^{-1}h^{-1}. \] On the other hand, the inverse of a product always satisfies \[ (gh)^{-1} = h^{-1}g^{-1}. \] Thus \[ g^{-1}h^{-1}=h^{-1}g^{-1}. \] Taking inverses gives \[ gh=hg. \]
6. No nonidentity element of \(G\) is equal to its inverse.
If \[ g=g^{-1}, \] then step 4 gives \[ \varphi(g)=g. \] The only fixed point of \(\varphi\) is the identity, so \(g=e\).
7. The order of \(G\) is odd.
By step 6, the inversion map pairs every nonidentity element \(g\) with a distinct element \(g^{-1}\). Thus \[ G\sm\{e\} \] is a disjoint union of two-element sets \[ \{g,g^{-1}\}. \] Hence \[ \abs G-1 \] is even, so \(\abs G\) is odd.
8. Q.E.D.
Step 3 proves part (1), while steps 4, 5, and 7 prove all claims in part (2).