Multiplicative Inverses in Non-Commutative Rings

My abstract book defines inverses (units) as solutions to the equation $ax=1$ then stipulates in the definition that $xa=1=ax$, even in non-commutative rings. But I'm having trouble understanding why this would be true in the generic case.

Can someone help me understand?

Book is "Abstract Algebra : An Introduction - Third Edition" By Thomas W. Hungerford. ISBN-13: 978-1-111-56962-4 (Chapter 3.2)

Edit:

Ok, I think I got it. Left inverses are not necessarily also right inverses. However, if an element has a left inverse and a right inverse, then those inverses are equal:

$$lx = 1 \\ xr = 1\\ lxr = r \\ lxr = l \\ r = l$$

Also, I assumed that 'unit' was synonymous with 'has a multiplicative inverse'. It is not; a 'unit' is an element that has both a left and right inverse, not just an inverse in general.

-

migrated from mathoverflow.netOct 28 '13 at 7:46

This question came from our site for professional mathematicians.

This is false. You can find a counterexample in, say, the ring of linear operators on an infinite-dimensional vector space. –  Qiaochu Yuan Oct 28 '13 at 7:32
(Every time you mention a book, give a complete reference)) –  Mariano Suárez-Alvarez Oct 28 '13 at 7:41
I may be missing something but a ring (with identity) has a group structure with respect to multiplication and addition and the multiplicative and additive inverses are unique. –  Mustafa Said Oct 28 '13 at 7:44
@MustafaSaid a ring does NOT have a group structure with respect to multiplication. consider $\mathbb{Z}/4\mathbb{Z}$, where $2x=2$ has more than one solution –  TBrendle Oct 28 '13 at 7:58
Doesn't "stipulates" imply that Hungerford is including $ax = xa =1$ as part of his definition? –  Robert Lewis Oct 28 '13 at 8:04

No, I have the book in front of me and he defines units as elements that have both a left and a right inverse, so $a$ is a unit if there exist elements $x$ and $y$ in $R$ such that $ax=ya=1$. Note $x$ need not equal $y$.
In his very next Remark he then proves that in this situation $x=y$, as a theorem, not as part of the definition.