# show that the composition of two involutions on R is an automorphism

Let R be a ring. An involution on R is a function $$\alpha : R \rightarrow R$$ such that, for all $$r_i ∈ R$$, we have $$\alpha(r_1 + r_2) = \alpha(r_1) + \alpha(r_2), \alpha(r_1r_2) = \alpha(r_2)\alpha(r_1)$$ and $$\alpha(\alpha(r_1)) = r_1$$

I guess my question is what does two involutions look like? Is this line of thought correct:

$$\alpha(R)\circ \alpha(R) = \alpha(\alpha(R)) = R$$, where $$\alpha(R)$$ is the involution $$\alpha(\alpha(R)) = R$$.

From here show the composition is one-to-one and onto to imply we have an automorphism.

The issue here is that $$\alpha$$ is an antiautomorphism, so it reverses the order of multiplication. If the ring is not commutative, it will not be a homomorphism of rings. However, if you have a second involution $$\beta$$ then $$\alpha\circ\beta$$ is still an additive homomorphism, and $$\alpha(\beta(r_1r_2))=\alpha(\beta(r_2)\beta(r_1))=\alpha(\beta(r_1))\alpha(\beta(r_2))$$ So $$\alpha\circ \beta$$ is a homomorphism, preserving the order of multiplication. Since both functions are invertible, so is the composition, so it is an automorphism.