TITLE: Shoe-socks property
QUESTION [0 upvotes]: 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?

REPLY [1 votes]: 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 $b^{-1}a^{-1}$.
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