Why Is a Group with Identical Elements Considered Unfaithful?

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SUMMARY

A group representation where ##D(e)=1## and ##D(s)=1##, with ##e \neq s##, is classified as unfaithful due to the lack of isomorphism. This situation arises when the group is denoted as ##(\{1,1\},\cdot)##, leading to confusion regarding the treatment of the set as having two elements despite containing only the identity element. The critical aspect is that both representations yield the same value, ##D(e) = D(s)##, indicating a failure to distinguish between distinct group elements.

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LagrangeEuler
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If I have some group representation ##D(e)=1##, ##D(s)=1## where ##e\neq s## it is called unfaithfull because it is not isomorphism.
If I denote this group by ##(\{1,1\},\cdot)##. My question is how I treat this set as a two element one, when I have only one element in the set? I'm a bit confused with this.
 
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It's not a two element set. The group only has one element, which is the identity. The point is that D(e) = D(s).
 

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