Proof. Let $G$ be a group, and $x,y\in G$. Then, we have $xy(xy)^{-1}=e$, where $e$ is the identity element of $G$. Multiplying on the left by $x^{-1}$, we get $y(xy)^{-1}=x^{-1}$. Multiplying once again on the left by $y^{-1}$ gives $(xy)^{-1}=y^{-1}x^{-1}$, which was what was to be shown.
Obviously this comes to the correct conclusion, but is it valid by the laws of algebra? The assumption that $xy(xy)^{-1}=e$ is directly from the fact that $G$ is a group (closed under inversion and the operation), but perhaps it's just my tendency to view $xy$ as two completely separate elements rather than an actual member of $G$ that's distracting me?
Is there a better way to show this? Because if it is valid, it seems (to me, at least, and beauty of the proof is in the eye of the writer..) more "intuitive" than the proof my book gave.
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$\begingroup$Another way to see this: $(xy)(y^{-1}x^{-1}) = x(yy^{-1})x^{-1}$ using associativity, and this equals $xex^{-1} = xx^{-1} = e$. Also show that $(y^{-1}x^{-1})(xy) = e$ in a similar way. Then use the fact that inverses are unique.
$\endgroup$ 2 $\begingroup$Yes, this is valid.
Whether it's more intuitive than the usual proof depends a bit on whether you view a inversion as an operation that comes with the group (i.e. a group is a set with a binary operation and an identity element and a unary operation, which satisfy such-and-such identities), or just as a fact that every group has to satisfy (a group is just a set with a binary operation and axioms such as $\exists e\forall z:e*z=z$ and $\forall x\exists y\forall z: x*y*z=z$).
In the former case your reasoning is probably the most straightforward; in the latter case it seems to be more direct to say that $y^{-1}x^{-1}$ is the inverse of $xy$ because it satisfies the defining property of inverses: $$ (xy)(y^{-1}x^{-1})=x(yy^{-1})x^{-1} = xex^{-1} = xx^{-1} = e $$
$\endgroup$ $\begingroup$Your proof is valid. Commonly this property is known as the "socks-shoes" property. If you consider $x$ as putting socks and $y$ as putting shoes, and $x^{-1}$ as takining off socks and $y^{-1}$ as taking of shoes, if $xy$ would represent putting on socks followed by shoes. In order to take them off, they must be removed in reverse order, that is, $y^{−1}x^{−1}$, you can see the identity as an analogy of wearing shoes-socks in daily life.
$\endgroup$ 6 $\begingroup$Well, when you ask if this is valid under rules of algebra, presumably you mean provable from group axioms.
The only subtlety is $x = y\implies zx = zy$, but this is trivial if we note that $(z\,\cdot)\,\colon G\to G$ is a function.
Now,
\begin{array}{c r l r} & (xy)(xy)^{-1}\!\!\!\!\! &= e &\text{($xy\in G\implies (\exists z\in G)\ z(xy) = (xy)z = e$)}\\ \implies & x^{-1}((xy)(xy)^{-1})\!\!\!\!\! &= x^{-1}e\\ \implies & ((x^{-1}x)y)(xy)^{-1}\!\!\!\!\! &= x^{-1}&\text{(associativity and neutral)}\\ \implies & (ey)(xy)^{-1}\!\!\!\!\! &= x^{-1}&\text{(inverse)}\\ \implies & y(xy)^{-1}\!\!\!\!\! &= x^{-1}&\text{(neutral)}\\ \implies & y^{-1}(y(xy)^{-1})\!\!\!\!\! &= y^{-1}x^{-1}\\ & &\ \vdots\\ \implies & (xy)^{-1}\!\!\!\!\! &= y^{-1}x^{-1} \end{array}
Thus, your proof works just fine within group theory.
But, the problem with this proof is that it doesn't generalize well to situations where not necessarily every element has inverse. For example, think of matrices, let's say that we have two invertible matrices $A$ and $B$. It is not a priori that $AB$ is invertible, as in case of groups. Thus, we need to prove it "manually" by guessing it's inverse. Not only matrices, think of invertible functions and composition of invertible functions, etc. All these situations can be handled as mentioned in other answers: prove $(xy)(y^{-1}x^{-1}) = e$ and use uniqueness of inverse if it exists. To be more precise, this works within scope of monoids, where there is neutral, but not necessarily inverse.
Let us show that inverse is unique in any monoid. Let $x\in M$, and let $y,z$ be it's inverses, i.e. $xy =yx = e$ and $xz = zx = e$, where $e\in M$ is neutral. Then, $$ y = ye = y(xz) = (yx)z = ez = z.$$
Finally, let $x,y\in M$ be invertible. We will show that $xy$ is invertible:
$$(xy)(y^{-1}x^{-1}) = x(y(y^{-1}x^{-1})) = x((yy^{-1})x^{-1}) = x(ex^{-1}) = xx^{-1} = e$$ and analogously $$ (y^{-1}x^{-1})(xy) = e,$$ so, by uniqueness $(xy)^{-1} = y^{-1}x^{-1}$
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