Basic invertible matrix theorem proof.

[Evil Robot]

If A and B are invertible matrices, can anyone prove that

(AB)^(-1) = B^(-1)*A^(-1) ?

This is not exactly the problem, I have, but my group theory problem is isomorphic to it :).

 

[Galileo]

Check:
With that expression for $(AB)^{-1}$, does the following hold?

$$(AB)^{-1}(AB)=1$$

 

[wais]

Yes, the basic invertible matrix theorem states that if A and B are invertible matrices, then the product AB is also invertible and its inverse is equal to the product of the inverses of A and B, i.e. (AB)^(-1) = B^(-1)*A^(-1). This can be proven using the definition of an inverse matrix and some basic properties of matrix multiplication.

Proof:

Let A and B be invertible matrices. This means that there exist matrices C and D such that AC = CA = I and BD = DB = I. We want to show that (AB)^(-1) = B^(-1)*A^(-1).

First, note that (AB)(B^(-1)*A^(-1)) = A(BB^(-1))A^(-1) = AIA^(-1) = AA^(-1) = I. Similarly, (B^(-1)*A^(-1))(AB) = B^(-1)(AA^(-1))B = B^(-1)IB = B^(-1)B = I.

Therefore, (B^(-1)*A^(-1)) is the inverse of AB, which means that (AB)^(-1) = B^(-1)*A^(-1). This completes the proof.

In group theory, a similar concept can be applied. If G and H are isomorphic groups, then the product GH is also isomorphic to the product of the isomorphic groups H and G. This can be proven using the definition of isomorphism and some basic properties of group multiplication.

Proof:

Let G and H be isomorphic groups. This means that there exists an isomorphism f: G → H. We want to show that GH is isomorphic to HG.

First, note that for any g1, g2 ∈ G, f(g1g2) = f(g1)f(g2) (since f is an isomorphism). Similarly, for any h1, h2 ∈ H, f(h1h2) = f(h1)f(h2).

Therefore, for any gh ∈ GH, we have f(gh) = f(g)f(h) ∈ HG. Similarly, for any hg ∈ HG, we have f(hg) = f(h)f(g) ∈ GH.

This shows that the map f': GH → HG defined by f'(gh) = f