- #1
Bashyboy
- 1,421
- 5
The problem is to verify ##(g^n)^{-1} = g^{-n}## is true ##\forall n \in \mathbb{Z}##. Here is my proof:
## (g^n)^{-1} = (\underbrace{g \star g~ \star ~...~ \star g}_{n~ \mbox{copies}})^{-1} \iff##
##(g^n)^{-1} = [(g \star ~...~ \star g) \star g]^{-1}##
Using ##(a \star b)^{-1} = b^{-1} \star a^{-1}## (1), which I have already proven, gives
##(g^n)^{-1} = g^{-1} \star (\underbrace{g \star g~ \star ~...~ \star g}_{n-1~ \mbox{copies}})^{-1} ##
If I apply (1) ##n-1## more times, I will arrive at
##(g^n)^{-1} = \underbrace{g^{-1} \star ~...~\star g^{-1}}_{n ~\mbox{copies}}##
The right hand side is by definition ##g^{-n}##.
What is wrong with this proof? My professor said it is wrong.
## (g^n)^{-1} = (\underbrace{g \star g~ \star ~...~ \star g}_{n~ \mbox{copies}})^{-1} \iff##
##(g^n)^{-1} = [(g \star ~...~ \star g) \star g]^{-1}##
Using ##(a \star b)^{-1} = b^{-1} \star a^{-1}## (1), which I have already proven, gives
##(g^n)^{-1} = g^{-1} \star (\underbrace{g \star g~ \star ~...~ \star g}_{n-1~ \mbox{copies}})^{-1} ##
If I apply (1) ##n-1## more times, I will arrive at
##(g^n)^{-1} = \underbrace{g^{-1} \star ~...~\star g^{-1}}_{n ~\mbox{copies}}##
The right hand side is by definition ##g^{-n}##.
What is wrong with this proof? My professor said it is wrong.
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