Prove that it is impossible that there exists $a,b,c\in G$ such that $a\neq b$ but $a\circ c = b\circ c$.

Asked Sep 14 '22 at 02:16

Active Sep 14 '22 at 02:16

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How would I do this without starting with the assumption that a = b? I looked around and I can't quite find a way to start this.

asked Sep 14 '22 at 02:16

1

As a hint - $c^{-1}$ exists. – CyclotomicField Sep 14 '22 at 02:18
But you may start with the assumption that $a\circ c = b\circ c$. – peterwhy Sep 14 '22 at 02:19
Prove that if $a \circ c = b \circ c$, then $a=b$. Also, you forgot to say that $(G,\circ)$ is a group. – azif00 Sep 14 '22 at 02:19
yes, usually the little circle is used for function composition. – qwr Sep 14 '22 at 02:32
Assume: a\neq b
then I apply the inverse of c to both sides and arrive at the equality nonetheless, thus a must equal b.

Would that be a sound proof?

Sorry for omitting G being a group.
– TruthSeekerWolfram Sep 14 '22 at 03:12
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You must say "I $\color{red}{\text{right}}$-multiply by $c^{-1}$. – Jean Marie Sep 14 '22 at 04:25

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