4
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Seems good to me $\endgroup$ – Kevin Quirin Oct 26 '15 at 14:52
  • 4
    $\begingroup$ 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
1
$\begingroup$

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 $e$.

$\endgroup$
  • 2
    $\begingroup$ 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
-1
$\begingroup$

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 $\ ab = e \Rightarrow ba = e \Rightarrow ab = ba$.

$\endgroup$
  • $\begingroup$ 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

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.