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I am studying groups and I had the question whether the implication in the title can be obtained or if, on the contrary, it is false.

My attempt to prove it was as follows:

Let $ a $ be an element of $ G $, define $ b = a^{-1} $, and suppose that $ ab = 1 $. Then we must show that $ ba = 1 $.

Since $ ab = 1 $, we have that $ b = abb $ and $ a = aba $, so

$$ ba = (abb)(aba) = (ab)(ba)(ba) = (ba)(ba). $$

The next step would be to see if in a group, the condition $ xx = 1 $ implies that $ x = 1 $.

I would like to know if the approach of the proof is correct and if you have any suggestions on how to show this last statement. Thank you for reading :D

Van Kampen
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    $(-1)^2=1$ but $-1\ne1$. Instead note $(ba)b=b(ab)=b$ and right-multiply by $b$ – J. W. Tanner Oct 10 '24 at 16:45
  • The condition $x^2=1$ does not imply $x=1$. Your question lacks sufficient details. What are your assumptions about the group? Are you assuming every element has a two-sided inverse? Are you assuming it has a two-sided identity? Are you only assuming the existence of a right identity and a right inverse to each element? – Arturo Magidin Oct 10 '24 at 16:48
  • What is true is that $xx=x$ implies $x=1$. So try $(ba)(ba) = b(ab)a$. – Arturo Magidin Oct 10 '24 at 16:50
  • This is true about groups. But if you start out with only certain axioms, then try to deduce this, it may require some effort. – GEdgar Oct 10 '24 at 16:50
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    Aren't all answers "yes" in a group, almost by definition, @ArturoMagidin? – Kan't Oct 10 '24 at 16:51
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    @Kan't Depends on how you define a group. You can define a group as a set (i) with an associative operation in which (ii) there is a two-sided identity and two-sided inverses for each element. Or (ii') there is a right identity and right inverses for each element; or (ii''') there is a left identity and left inverses for each element; or (ii'''') the equations $ax=b$ and $xa=b$ always have a unique solution; or (ii''''') the equations $ax=b$ and $xa=b$ always have at least one solution, etc. – Arturo Magidin Oct 10 '24 at 16:53
  • @Kan't While all definitions imply that if $ab=1$ then $ba=1$, how you prove that depends on your definition. If you assume (ii), then the proof is by showing that inverses are unique; If you assume (ii'), then the argument is a bit more involved, etc. – Arturo Magidin Oct 10 '24 at 16:54
  • The question arose from observing the group $ R^{\ast} $ of invertible elements in a ring. I have the definition that an element $ a $ is invertible in a ring $ R $ if there exists a $ c \in R $ such that $ ac = 1 $.

    I was testing the existence of inverses. If $ a \in R^{\ast} $, then there exists a $ b \in R $ such that $ ab = 1 $. I want to prove that $ b \in R^{\ast} $ by showing that $ ba = 1 $.

     – Van Kampen Oct 10 '24 at 16:57
  • Your definition of "invertible" in a ring is incorrect. What you have defined is a right invertible element, and the result you want is false, as is well-known. For example, the ring of linear transformations of the vector space of all real sequences contains the left-shift operator $\lambda (a_0,a_1,\ldots)= (a_1,a_2,\ldots)$, and the right-shift operator $\rho(a_0,a_1,\ldots) = (0,a_0,a_1,\ldots)$. Then $\lambda\rho=1$ but $\rho\lambda\neq 1$. The set of right-invertible elements need not be a group, because $c$ need not be right-invertible. – Arturo Magidin Oct 10 '24 at 17:01
  • Oh, okay, I will review the definition again; I may have overlooked something. Thank you very much! :D – Van Kampen Oct 10 '24 at 17:04
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    You have either overlooked something, or you should throw away the source that gave you that definition of "invertible" in a ring that is not necessarily commutative. – Arturo Magidin Oct 10 '24 at 17:10
  • Any element of a group $a$ has a two sided inverse $a^{-1}$, so if $ab=1$, then $a^{-1}=a^{-1}1=a^{-1}(ab)=(a^{-1}a)b=1b=b$. – John Douma Oct 10 '24 at 19:00

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