Ytukyg

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

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

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

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.

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.$$

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

All the answers so far have shown that $(aast b)ast(b^{-1}ast a^{-1}) = (b^{-1}ast a^{-1})ast(aast b) = e$ so $(aast b)^{-1} = b^{-1}ast a^{-1}$. 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 $aast b$. Then we have

begin{align*}
(aast b)ast x &= e\
aast(bast x) &= e\
a^{-1}ast(aast(bast x)) &= a^{-1}ast e\
(a^{-1}ast a)ast(bast x) &= a^{-1}\
east(bast x) &= a^{-1}\
bast x &= a^{-1}\
b^{-1}ast(bast x) &= b^{-1}ast a^{-1}\
(b^{-1}ast b)ast x &= b^{-1}ast a^{-1}\
east 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, $(aast b)^{-1} = b^{-1}ast a^{-1}$.

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.