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prove map σ:y→xyx⁻¹ is bijective

 
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Mar15-12, 08:25 PM   #1
 

prove map σ:y→xyx⁻¹ is bijective


1. Let G be any group and x∈G. Let σ be the map σ:y→xyx⁻¹. Prove that this map is bijective.
It seems to be written strangely, since it never really says anywhere that y is in G, but I guess that must be an assumption.


2. bijective=injective+surjective.
in order to prove injective, we need to show that y1≠y2→xy1x⁻¹≠xy2x⁻¹
and in order to prove surjective, we need to show that for every g in G, there exists a y in G such that xyx^-1=g.




3. I think that I can say: Let y=x^-1gx. Then xyx-1=g and we are done for surjective.
I don't really know how to "show" injective, since it seems obvious.
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Mar15-12, 09:19 PM   #2
 
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Instead of showing ##y_1\ne y_2 \to xy_1x^{-1}\ne xy_2x^{-1}## try showing the contrapositive.
Mar15-12, 09:28 PM   #3
 
So I show that xy1x-1=xy2x-1→y1=y2 by simply left-multiplyng both sides by x-1 and right-multiplying both sides by x? Is that too simple?

Also, does my thinking on surjective work?
Mar15-12, 09:32 PM   #4
 
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prove map σ:y→xyx⁻¹ is bijective


Yes, it all looks OK to me.
Mar15-12, 09:33 PM   #5
 
Ok, thanks!
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