Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 8

Let $n=\mathrm{ord}(a)$, which is finite. A group homomorphism preserves powers and the identity, soIf $r=\mathrm{ord}(f(a))$, divide $n$ by $r$: $n=qr+s$ with $0\le s<r$. Then $e^{\prime }=f(a{)}^n=f(a{)}^s$, and minimality of $r$ forces $s=0$. ThusIf $f$ is a group isomorphism, apply the same argument to its inverse: $\mathrm{ord}(a)\le \mathrm{ord}(f(a))$. The two inequalities yield

The two products are conjugate:Conjugation by $a^{-1}$ is a group automorphism, so part (a) provesDirectly, cancellation gives $(ba{)}^r=a^{-1}(ab{)}^{r}a$ for every positive integer $r$. Thus a power of one product is the identity exactly when the same power of the other is the identity. This proves that conjugation preserves element order, even without the finiteness assumption.

The hypothesis implies $g^2=e$ for every $g\in G$, including the identity, so $g^{-1}=g$. For arbitrary $a,b\in G$, their product also equals its own inverse. The inverse-of-a-product rule therefore givesThus $G$ is an abelian group. Finiteness is not needed for this implication; it holds for every group in which each element squares to the identity.