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Showing the set of units $R^{x}$ of a ring with identity $R$ is closed under multiplication.

I am trying to show that the set $R^{x}=\left \{a\in R: \exists b \in R : ab=1 \right \}$ is a group under multiplication. I am stuck at the closure under multiplication part. Here is my work:

Let $a,b\in R^{x}$. Then there exist $c,d\in R$ such that $ac=1$ and $bd=1$.

I thought maybe multiplication will help.

$1 = (1)(1) = acbd$

But it is not assumed that $R$ is abelian.

Any ideas what I should do?

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2 Answers 2

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Hint: recall what the general expression for the inverse of a product for any group is. That should give you a suitable candidate.

Solution: in any group $G$, we have $(xy)^{-1} = y^{-1}x^{-1}$ for $x,y \in G$. This motivates the candidate $dc$. In effect, $(ab)(dc) = a(bd)c = a(1)c = ac = 1$, and so $ab \in R^\times$.

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If $ac=1$ and $bd=1$ then $(ab)(dc)=a(bd)c=ac=1$.

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