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- Let G be a Group.
- 1.
![{\displaystyle \forall \;g,a,b\in G:(g\ast a=g\ast b)\rightarrow (a=b)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/adbb82f3697f3eaf29207b9c5c8b8b1f789b7f02)
- 2.
![{\displaystyle \forall \;g,a,b\in G:(a\ast g=b\ast g)\rightarrow (a=b)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc73c3913e8e28dd24a33d959e781134eaab3fa8)
0. Choose such that
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1.
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definition of inverse of g in G (usage 1)
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2.
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0.
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3.
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is associative in G
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4.
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g-1 is inverse of g (usage 3)
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5.
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eG is identity of G(usage 3)
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if a*g = b*g...
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a = a*g*g-1
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b*g*g-1 = b
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then a = b.
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- if a, b, x are in the same group, and x*a = x*b, then a = b
- a, b, and g have to be all in the same group.
has to be the binary operator of the group.
- G has to be a group.