Proving Surjectivity of Mapping A_g:G ---> G for Automorphism

  • Thread starter Thread starter RVP91
  • Start date Start date
  • Tags Tags
    Surjective
RVP91
Messages
50
Reaction score
0
I have a mapping A_g:G ---> G defined by
A_g(x) = g^-1(x)g (for all x in G)

and as part of showing it is an automorphism i need show it is surjective.

I'm not entirely sure how to do this but have made an attempt and would appreciate and feedback or hints to what I actually need to show. I know the definition of surjectivity and also that a mapping is surjective iff Im(of mapping) = G

My attempt:
It is surjective since if x is in G, then gxg^-1 is in G and then A_g(gxg^-1) = (g^-1)gx(g^-1)g = idxid = x.

Thanks in advance
 
Mathematics news on Phys.org
Yep, surjective just means that every element has a pre-image, and you have shown that by writing down the pre-image explicitly.
 
thanks
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top