Suppose that instead of the property $ab=ba=e$ a group G has the condition that for every element $a$ there exists an element $b$, such that $ab=e$. Prove that $ba=e$. Is the following a valid proof?

Since $ab=e$ then under the condition of the group there exists an element $k$ such that $bk=e$ for some $k$ in the group.

Now $bk=e$ so $abk=ae$ therefore $(ab)k=a$ and finally $ek=a$ and $k=a$.

Is this a valid proof?

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    $\begingroup$ Seems good to me $\endgroup$ – Kevin Quirin Oct 26 '15 at 14:52
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    $\begingroup$ Wait a moment, if a group $G$ satisfies the condition: $$\forall a,b\in G \Rightarrow ab = \mathrm{id}_G,$$ then it is forced to be the trivial group $$G = \{\mathrm{id}_G\}.$$ Which condition are you exactly imposing on your group? $\endgroup$ – Giovanni De Gaetano Oct 26 '15 at 15:03
  • $\begingroup$ I assume the first sentence in this post is a misinterpretation of the problem to be solved. $\endgroup$ – David Hill Oct 26 '15 at 15:07
  • $\begingroup$ @GiovanniDeGaetano Yes I corrected it. Thank you for pointing that out... $\endgroup$ – Stefan Oct 26 '15 at 15:32
  • $\begingroup$ good to go .... $\endgroup$ – Kushal Bhuyan Oct 26 '15 at 15:47

I assume that you mean if for SOME $a,b$ (not every) $ab = e$ then $ba = e$. Your proof is valid, but you could write it much easier without playing with $k$. $$ab = e \Rightarrow bab = b.$$ If you already know the cancellation law, then we are done. Otherwise you may continue by writing $baba = ba$, so that $(ba)^2 =ba$. Just note that the only idempotent in a group is $e$.

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    $\begingroup$ The cancellation law is a consequence of the existence of unique inverses (as is the fact that $e$ is the only idempotent). Without expanding the proofs of one of those statements, this is just hiding the desired statement rather than proving it. $\endgroup$ – Milo Brandt Oct 26 '15 at 15:45
  • $\begingroup$ @MiloBrandt Cancellation law has nothing to do with "uniqueness". Ones you have a right (left) inverse then you have the right (left) cancellation law. In this problem we have $G$ is a group, so by definition we have right and left inverses. $\endgroup$ – Amin Oct 26 '15 at 16:13
  • $\begingroup$ I think after: $$ab = e \Rightarrow bab = b.$$ We can probably just conclude by associativity $(ba)b=b$, and since the identity is unique then $ba=e$ from there? $\endgroup$ – Stefan Oct 26 '15 at 17:15
  • $\begingroup$ @StefanT. You should be careful here. As Milo mentioned, using uniqueness is not allowed here unless you prove it. Here you can write $(ba)b = eb$ and use the right cancellation. $\endgroup$ – Amin Oct 26 '15 at 17:46

Theorem $(A)$: For every $\{a, b\} \subset \mathbb{R}, \ ab = c \ $ and $\ a \neq 0$ $$\implies b = \frac{c}{a}$$ Proof: $$\begin{align} ab &= c \\ \iff \bigg(\frac{1}{a}\bigg)(ab) &= \bigg(\frac{1}{a}\bigg)c \\ \iff \bigg(\frac{1}{a} \times a\bigg)b &= \bigg(\frac{1}{a}\bigg)c \\ \iff 1\times b &= \bigg(\frac{1}{a}\bigg)c \\ \iff b &= \bigg(\frac{1}{a}\bigg)c \\ \iff b &= \frac{c}{a} \end{align}$$

$(A)$ is derived from the multiplicative cancellation law, which proves that for every $\{a, b, c\} \subset \mathbb{R}, \ ac = bc \ $ and $\ c \neq 0 \Rightarrow a = b$. So now all you have to do is substitute $c = e$ and $b = a$ in the proof of $(A)$ and you can prove that $\ ab = e \Rightarrow ba = e \Rightarrow ab = ba$.

  • $\begingroup$ I got a negative vote, so what can I do to improve my answer? Edit: Sorry, I forgot to add in "substitute $a = b$" as well as substituting $c = e$. But it is fixed now $\endgroup$ – Mr Pie Oct 19 '17 at 2:43

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