Application of associative property in the proof of finding the solution of the equation ax=b.

Question

I am trying to find the solution for the equation $ax=b$ where $a,b \in G$ and $G$ is a group with respect to the operation, multiplication. This is what I gathered from different books $$ax=b$$ pre multiply both sides with $a^{-1}$. $$(a^{-1})ax=(a^{-1})b$$ Then it says according to associative property $aa^{-1}=e$. $$ex=(a^{-1})b$$ $$x=(a^{-1})b.$$ But the thing that I don't understand is the step where the associative property is applied above because according to associative property(w.r.t multiplication)

$$ \forall a,b,c \in G , a(bc)=(ab)c$$ and in this case we know that $a$ and $a^{-1}$ belongs to $G$ but $x$ does not. So how can one apply associative property in such a situation. Please elucidate. Thanks in advance.

Answer 1

I think your confusion about whether $x$ lies in the group was addressed in the comments.

The associativity step that you're looking for comes right after you left-multiply by $a^{-1}$ to get $$(a^{-1})(ax) = a^{-1}b.$$ On the left you need to apply associativity to see that: $$a^{-1}(ax) = (a^{-1} a)x = ex.$$

Hope this helps!

Answer 2

So you have $ax = b$. They are both different ways of writing the same element. I will refer to that element as $THEELEMENT$. We have $b =THEELEMENT$ and $ax = THEELMENT$.

Now if we multiply $THEELEMENT$ by $a^{-1}$ we get a second element $a^{-1}\cdot THEELEMENT$. I will refer to this as $THEOTHERELEMENT$.

So $THEOTHERELEMENT = a^{-1}\cdot THEELEMENT = a^{-1}\cdot (ax) = a^{-1}\cdot b$

Now the text is going to make the claim that

$THEOTHERELEMENT = a^{-1}\cdot (ax) = (a^{-1}a)\cdot x$.

But why can we say that if do $ax$ first to get $THEELEMENT$ and then do $a^{-1}THEELEMENT$ that that will be the same as doing $(a^{-1}a)$ first and then doing $(a^{-1}a)\cdot x$ to get the same thing?

Well, THAT is where associativity comes in.

$a^{-1}(ax) = (a^{-1}a) x$ by associativity

And so

$a^{-1}(ax) = a^{-1}b$

$(a^{-1}a)x = a^{-1}b$

$ex = a^{-1}b$

$x = a^{-1}b$.

The thing is this was so intuitively obvious you didn't even bother to write down your terms with parenthesis to tell us what order we were doing them in. Why not? Because you didn't think it mattered. And why didn't you think it mattered? Because it doesn't because of associativity.

But if you didn't have associativity it would matter and would have to write $a^{-1}(ax) =a^{-1}b$. Writing $(a^{-1})ax$ would both be unclear because you didn't write the order, and wrong if you assume tthe order is left to right$.