Showing this is surjective

  • Thread starter RVP91
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  • #1
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Main Question or Discussion Point

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
 

Answers and Replies

  • #2
CompuChip
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Yep, surjective just means that every element has a pre-image, and you have shown that by writing down the pre-image explicitly.
 
  • #3
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thanks
 

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