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prove map σ:y→xyx⁻¹ is bijective |
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| Mar15-12, 08:25 PM | #1 |
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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. |
| 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.
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| Mar15-12, 09:28 PM | #3 |
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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.
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| Mar15-12, 09:33 PM | #5 |
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Ok, thanks!
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