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A cute name for an elementary result in group theory.
We let a-1 denote the inverse of an element a in a group, ie the the element such that a-1*a = a*a-1 = e, where e is the identity element. Then (using the associative property of the operation in a group)

(a*b)*(b-1*a-1) = a*(b*b-1)*a-1 = a*a-1 = e

Thus the inverse of a*b is b-1*a-1.
This result is sometimes called the shoes-and-socks theorem because of the following analogy: if you want to undo the act of putting your socks and then your shoes on, you first have to remove your shoes and then take off your socks.

Despite the funny name this is a relatively important finding from linear algebra*, and it is exceedingly simple to use and prove. It takes two square matrices A and B, which have n rows and n columns. If these two matrices are invertible, i.e. their rank = n, then we know this about the inverse of their product: (AB)-1 = B-1A-1.

This finding is referred to as the shoe-sock theorem because it resembles the steps needed to invert (that is, undo) the action of putting on socks and shoes. If A is for socks, B is for shoes, and -1 is the process of taking them off, to take them both off (AB)-1 one must first take off the shoes B-1, then the socks A-1. Hey, I said it was a funny name, not a particularly insightful one...

One proof is nice and easy, though I'm sure harder ones can be devised without much difficulty. This proof relies on the fact that we need to prove B-1A-1 to be the inverse of AB. We'll do this by showing

(B-1A-1)(AB) = In

and

(AB)(B-1A-1) = In.

In other words, we'll show that multiplying AB on either side will produce the identity matrix of size n (that is, In), which is necessary and sufficient proof that the inverse of AB is B-1A-1.

First: (B-1A-1)(AB) = B-1(A-1A)B = B-1InB = B-1B = In -- cool.
And also: (AB)(B-1A-1) = A(BB-1)A-1 = AInA-1 = AA-1 = In -- very cool, statement proven.

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*User zfcorey reminded me that this rule is true of any abstract algebra with an identity.

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