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I am supposed to prove that given a commutative ring $R$, the set of units $R^{\times}$ is a group. I checked the axioms of a group and it all came down to noting that if $a,b\in R^{\times}$, then $ab\in R^{\times}$. This is just saying that the product of two units is a unit and that is fine with me.

What I am confused about is the statement that the ring is commutative. It seems to me that the set of units in any ring (with identity) is a group. Is that true?

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    $\begingroup$ You are correct, $R$ need not be commutative; they're just trying to make your proofs a bit easier. $\endgroup$ – vadim123 Apr 21 '15 at 19:17
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The proof is this:

Let $U$ be the set of unit of your ring $R$. Let's suppose that $1 \in R$. Then:

1) It exists the neutral element;

2) If $a,b \in U$, then $a*b \in U$, because $(a*b)^{-1}=b^{-1}*a^{-1}$; in fact:

$$(a*b)*(b^{-1}*a^{-1})=a*(b*b^{-1})*a^{-1}=a*a^{-1}=1$$

3) The associativity follows from the definition of $R$ as a ring.

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