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Prove map σ:y→xyx⁻ is bijective

by catherinenanc
Tags: bijective, prove
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catherinenanc
#1
Mar15-12, 08:25 PM
P: 49
1. Let G be any group and x∈G. Let σ be the map σ:y→xyx⁻. 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⁻≠xy2x⁻
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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LCKurtz
#2
Mar15-12, 09:19 PM
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Instead of showing ##y_1\ne y_2 \to xy_1x^{-1}\ne xy_2x^{-1}## try showing the contrapositive.
catherinenanc
#3
Mar15-12, 09:28 PM
P: 49
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?

LCKurtz
#4
Mar15-12, 09:32 PM
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Prove map σ:y→xyx⁻ is bijective

Yes, it all looks OK to me.
catherinenanc
#5
Mar15-12, 09:33 PM
P: 49
Ok, thanks!


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