I am having trouble with the flip and switch for inverses:

$$(gh)^{-1}=h^{-1}g^{-1}$$

Question

I am having trouble with the flip and switch for inverses:

$$(gh)^{-1}=h^{-1}g^{-1}$$

anonymous · Answer

why does this rule apply like this

amistre64 · Answer

is that functions or just variables?

anonymous · Answer

i'm trying to prove that g,h are elements of an abelian group G that forms a subgroup H

amistre64 · Answer

sounds like abstract algebra, i read a book a while back but cant recall the details

anonymous · Answer

that inverse law is used to show that is it closed under the binary operation

amistre64 · Answer

abelian being... commutative?

anonymous · Answer

yeah, then there are 5 requirements for a subgroup must be closed... elements must have inverses, contain an identity etc...... I was just curious to why the inverse rule works as it does

amistre64 · Answer

the inverse states that there is some element such that when applied it equals an identity ... if im not mistaken

amistre64 · Answer

its rather self explanatory for numbers; but they get bent when doing abstract

phi · Answer

Is this convincing?
given $(AB)^{-1} (AB) = I $
also,we have $B^{-1}A^{-1}A B =I $
so $(AB)^{-1} (AB)= B^{-1}A^{-1}(A B) $
and $(AB)^{-1} = B^{-1}A^{-1} $