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Theorem: Let $G$ be a group with operation $\star$. For all $a,b,c\in G$, if $a\star b=a\star c$, then $b=c$.

I've got a proof, but I'm not sure it is correct (I'm not sure that I can apply the operation $\star$ on both sides):

Proof: Every element in $G$ has an inverse, so let $a^{-1}$ be the inverse of $a$. Then we can say $$a^{-1}\star a\star b=a^{-1}\star a\star c.$$ But $a\star a^{-1}=e$, the identity element of $G$. Then $$e\star b=e\star c,$$ and since $e$ is the identity element, $b=c$.

amWhy
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Caleb Jares
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    In a first encoounter with group theory, it's a good idea to justify everything explicitly. Here, you have used associativity, and you should note that. – Gerry Myerson Feb 27 '13 at 01:53
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    In general, you can do anything to one side of an equation, provided you do the same thing to the other side. – Gerry Myerson Feb 27 '13 at 01:54
  • Although sometimes it is possible to do something to both sides of an equation, and obtain a new equation that is logically weaker than the previous one (e.g. $a=b\implies a^2=b^2$ but $a^2=b^2\not\Rightarrow a=b$). – anon Feb 27 '13 at 02:44
  • @anon, going from $a^2=b^2$ to $a=b$ is not doing the same thing to both sides. – Gerry Myerson Feb 28 '13 at 05:49
  • @GerryMyerson Isn't going from $a^2=b^2$ to $a=b$ the same as $a^2=b^2\implies a=b^2/a$, and because $a=b$, $\implies a=b^2/b=b$. – Caleb Jares Feb 28 '13 at 17:41
  • @Gerry I never said anything about "going from $a^2=b^2$ to $a=b$;" the order of logical implications is simply to illustrate that the statement $a=b$ is stronger than $a^2=b^2$. The "going from blah to blah" part is simply squaring. [Caleb: no.] – anon Feb 28 '13 at 20:43
  • Caleb, you can't assume $a=b$ when trying to deduce $a=b$. – Gerry Myerson Feb 28 '13 at 21:32
  • @GerryMyerson I meant that if you went from $a=b$ to $a^2=b^2$, you can don't lose any information because you still have the two equations. – Caleb Jares Feb 28 '13 at 22:42

1 Answers1

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You did just fine! No problems there.

For the nitpicker, you may want to start off, also, with "Since $G$ is a group,

  1. $G$ contains a unique identity, which we'll denote $e$, such that for all $a \in G,\;\; a\star e = e \star a = a.$
  2. And it also follows that every element $a \in G\,$ has a unique inverse, which we'll denote $a^{-1}$, such that $a\star a^{-1} = a^{-1}\star a = e$.

Then everything follows precisely as you argued. But as Gerry Myerson pointed out, you may also want to add justification as below for using associativity:$$a \star b = a\star c \tag{hypothesis}$$ $$\iff a^{-1}\star (a\star b)= a^{-1}\star (a\star c) \quad\quad\tag{left "multiplication" by $a^{-1}$}$$ $$\iff (a^{-1} \star a) \star b = (a^{-1}\star a) \star c \quad\tag{$G$ is associative}$$ $$ \iff e\star b = e\star c \tag{2}$$ $$\iff b = c \tag{1}$$

amWhy
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  • And yes, as Gerry Myerson assured you, if $x$ is any element, you can apply $x \star$ to each side of the equation, provided it's on the left of each side of the equation, (or else apply $\star x$ provided it's to the right side of each equation). In this case, you proceeded precisely as desired. – amWhy Feb 27 '13 at 02:08
  • To elaborate on amWhy's comment, for each fixed element $g$ in the group $G$, the function $f: G \rightarrow G$ defined by $f(x) = gx$ is a function. (This can be proved by making use of the fact that the group operation is a function from $G \times G$ to $G.$) Therefore, the "algebraic version of the vertical line test" -- which says that if $a,$ $b \in G$ and $a=b,$ then $f(a) = f(b)$ -- justifies left-multiplying both sides of an equation by $g.$ Similarly for right-multiplying both sides of an equation by $g.$ – Dave L. Renfro Feb 27 '13 at 20:15