Shoe-socks property

Question

Is it true that for every group $(G,*)$ and any $a,b\in G$, $$(a*b)^{-1}=b^{-1}*a^{-1} \;\;?$$ Why, or why not?

Answer 1

$$(a*b)*(b^{-1}a^{-1}) = a*(b*b^{-1})*a^{-1} = a*e*a^{-1} = a*a^{-1} = e$$

And likewise $(b^{-1}a^{-1})*(a * b) = e$.

So with associativity of the operation in a group, and by the mere definitions of inverse elements in a group, $$(a*b)^{-1} = b^{-1}a^{-1}$$

Answer 2

This is true because $(a \ast b) \ast (b^{-1} \ast a^{-1}) = a \ast b \ast b^{-1} \ast a^{-1} = a \ast e \ast a^{-1} = a \ast a^{-1} = e$. We used that multiplication is associative.

Answer 3

It is true, because of associativity: $$(ab)(b^{-1}a^{-1})=a(bb^{-1})a^{-1}=a*e*a^{-1}=e \\(b^{-1}a^{-1})ab=b^{-1}(a^{-1}a)b=b^{-1}*e*b=e.$$

Answer 4

You claim that the inverse of $ab$ is $b^{-1}a^{-1}$. Then just multiply it and see what happens. Notice that inverses are unique in a group.

Answer 5

All the answers so far have shown that $(a\ast b)\ast(b^{-1}\ast a^{-1}) = (b^{-1}\ast a^{-1})\ast(a\ast b) = e$ so $(a\ast b)^{-1} = b^{-1}\ast a^{-1}$. Of course this is completely valid, but it may be hard to see where the expression $b^{-1}\ast a^{-1}$ came from.

Let $x$ be the inverse of $a\ast b$. Then we have

\begin{align*} (a\ast b)\ast x &= e\\ a\ast(b\ast x) &= e\\ a^{-1}\ast(a\ast(b\ast x)) &= a^{-1}\ast e\\ (a^{-1}\ast a)\ast(b\ast x) &= a^{-1}\\ e\ast(b\ast x) &= a^{-1}\\ b\ast x &= a^{-1}\\ b^{-1}\ast(b\ast x) &= b^{-1}\ast a^{-1}\\ (b^{-1}\ast b)\ast x &= b^{-1}\ast a^{-1}\\ e\ast x & = b^{-1}\ast a^{-1}\\ x &= b^{-1}\ast a^{-1}. \end{align*}

By uniqueness of the inverse element, or by verification as in the other answers, $(a\ast b)^{-1} = b^{-1}\ast a^{-1}$.

Answer 6

Suppose $a$ and $b$ are symmetries of some underlying object $X$. (All group elements can be understood in this way.) Then $ab$ is the composition of $a$ and $b$, which means that it is the transformation of $X$ that results when you first transform $X$ with transformation $a$, and then transform the result with transformation $b$.

Now suppose you would like to undo the transformation of $X$ that you have just done with $ab$. You must first undo $b$, then undo $a$. The transformation that undoes $b$ is exactly $b^{-1}$, and the transformation that undoes $a$ is $a^{-1}$. To undo $b$ and then $a$, in that order we compose these, in order, first $b^{-1}$ then $a^{-1}$. So the transformation that undoes $ab$ is $b^{-1}a^{-1}$.