mint river
is it right that i supposed that $ab = ca$ here? Here is my proof as follows:
Suppose $ab = ca$, thus $b = c$. Multiplying both sides by $a$ gives $ab = ac$. Since $ab = ca$ as well, it follows that $ac =ca $ for all $a$ and $c$, thus $G$ is abelian. Conversely, if $G$ is abelian and $ab = ca$, then $ab = ac$. Consequently, $b = c$ by cancellation.
fresh sedgeBOT
vast hamletBOT
herrperſon
swift citrus
is it right that i supposed that $ab = ca$ here? Here is my proof as follows:
Su...
There's one minor flaw in your proof
In the => direction, you have not shown that your group operation is commutative for every pair of elements. I know it looks like you did, but consider what your starting hypothesis was. You are only picking a,b,c that satisfy ab=ca. Your choices for a and c are not completely arbitrary here.
You need to instead start with some arbitrary a and c, and show that some b will exist that satisfies ab=ca
(Don't overthink it though, it's very easy to find this b value, but you do need to explicitly do it)
tepid snow
is it right that i supposed that $ab = ca$ here? Here is my proof as follows:
Su...
adding onto what SWR said, this is why it's very important to have quantifiers in your proof
as an example,
Claim: if $R$ is a transitive & symmetric binary relation, then it is reflexive (and hence an equivalence relation).
vast hamletBOT
LY
tepid snow
Proof: We know $xRy$ and $yRx$ since $R$ is symmetric, and by transitivity $xRy$ and $yRx \implies xRx$ so $R$ is reflexive.
vast hamletBOT
LY
mint river
In the => direction, you have not shown that your group operation is commutative...
So… suppose $a$ and $c$ are arbitrary elements of $G$. Then there exists $b = a^{-1}ca \in G$ s.t. $ab = aa^{-1}ca =ca$, therefore $ab = ca$?
vast hamletBOT
herrperſon
mint river
Can i do the same thing with… say arbitrary elements a and b or b and c and then find a c or a that satisfies ab = ca?
swift citrus
So… suppose $a$ and $c$ are arbitrary elements of $G$. Then there exists $b = a^...
There. That is perfectly satisfactory.
swift citrus
Can i do the same thing with… say arbitrary elements a and b or b and c and then...
you can find $b$ given $a$ and $c$, and you can find $c$ given $a$ and $b$, but I do not believe it is always possible to find $a$ given $b$ and $c$. Such an $a$ may not always exist.
vast hamletBOT
SWR
mint river
Okay
mint river
you can find $b$ given $a$ and $c$, and you can find $c$ given $a$ and $b$, but ...
…Wouldn't this only prove that for all a and c, there exists b such ab = ca, not that for all a, b, c ab = ca ( not for all b as well?)