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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.

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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.

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  • $\begingroup$ Can you explain the invertibility leading to automorphism, please? $\endgroup$ – user551155 Mar 30 '20 at 0:32
  • $\begingroup$ @user A homomorphism that is invertible is an automorphism. $\endgroup$ – Matt Samuel Mar 30 '20 at 0:33
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Hint: An antihomomorphism composed with an antihomomorphism is a homomorphism.

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