I am presented with the following task:

Show that if $U$ is the collection of all units in a ring $\langle R, +, \cdot\rangle$ with unity, then $\langle U, \cdot\rangle$ is a group.

I am still not confident with proofs in Abstract Algebra, and I'd like to receive some verification/slaughter/constructive critique on my reasoning here. I do not know if there are any mistakes, or how serious they are.

We know that there is an identity element, since the unity itself is a unit, and therefore in the group. It follows directly from the definition of a unit that there exists an inverse for every element in the group, and associativity follows directly from the Ring-axioms. As for closure, we need to show that the product of two units is also a unit. Let $a, b \in U$ be two elements with the property that $a\cdot b \ne e$, where $e$ is the identity element in $\langle U, \cdot \rangle$. We know that there exists elements $a', b' \in U$ with the property that $a' \cdot a = e = a' \cdot a$ and $b' \cdot b = e = b \cdot b'$. Let $a', b'$ be two such elements with the property that $a' \cdot b' \ne e$. We may now compute $(a\cdot b) \cdot (a' \cdot b')$, and from the Ring-axioms we have associativity under multiplication, so we may write the expression as $a \cdot (b \cdot a' \cdot b') = a \cdot (b \cdot b' \cdot a') = a \cdot (e \cdot a') = a \cdot a' = e$. Thus we have shown that the product of any two elements is a unity, which leads to closure. All the group-axioms are met and $\langle U, \cdot \rangle$ is a group.

I have to apologize if you find the argument needlessly long; I've been having some trouble with assuming things I am not supposed to assume (like what I'm going to prove), so I wanted to stay clear of that.

  • $\begingroup$ Do it with $(a.b).(b'.a')$ instead of $(a.b).(a'.b')$. Your proof is only valid if $R$ is commutative. Secondly I see no reason to demand that $a.b\ne e$. $\endgroup$ – drhab Mar 12 '14 at 19:40

This looks pretty good, assuming your ring is commutative (as noted in a comment above). Usually we discuss closure first, because without it, there's not really anything to talk about, but that's not essential.

What's required to establish is just: 0) The product of two units is a unit - you discovered that proof, and it will be fully general if you use $b'a'$ as your inverse instead of $a'b'$; 1) Multiplication is associative - free from the ring axioms, as you noted; 2) The multiplicative identity from the ring is a unit and serves as our group identity - got that; 3) The multiplicative inverse of a unit exists and is another unit - free from the definition of "unit", as you noted.

You established all of those things, so you're good. I'm sure the wording could be tightened up, but that will come with time. The only thing I'd really take issue with now is that the product argument is a little clunky. There's no need to assume that $ab\neq e$. You can really just say:

"Let $a$ and $b$ both be units, with multiplicative inverses $a'$ and $b'$. Then the element (b'a') is a multiplicative inverse for $ab$, so $ab$ is also a unit."

  • $\begingroup$ I fail to see where I assumed that $R$ is commutative. I (believe I) only used the commutative property $a\cdot a' = a'\cdot a$, which is a property of all inverses, commutative or not. $\endgroup$ – Andrew Thompson Mar 12 '14 at 19:50
  • 1
    $\begingroup$ It's where you went from $(b\cdot a'\cdot b')$ to $(b\cdot b'\cdot a')$. You swapped $a'$ and $b'$, but they aren't inverses of each other. $\endgroup$ – G Tony Jacobs Mar 12 '14 at 19:53
  • $\begingroup$ Of course. The statement is also true in the general case, I suppose, so I'll try to fix it up and make an edit by tomorrow. $\endgroup$ – Andrew Thompson Mar 12 '14 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.