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Suppose the set of residues $A=\{\bar 1, \bar 3, \bar 4, \bar 5, \bar 9\}$, $f(x,y)=x*y$, $\mathbb Z_{11}$ so the group is $G=\{A, f\}$ in $\mathbb Z_{11}$. The identity is $\bar 1$ as demonstrated by $\bar 3 *\bar 4=\bar{12} =\bar 1$. Suppose $\phi(x,y)=e^{x+y}$ where $e$ is the euler number to test the group homomorhism identity preserving theorem here. Now $B=\{e^{\bar 1}, e^{\bar 3}, e^{\bar 4}, e^{\bar 5}, e^{\bar 9}\}$. So the intended group is $H=\{B,\phi\}$ in $\mathbb Z_{e^{11}}$ but $e^{\bar 9+\bar 9}=e^{\bar {18}}=e^{\bar 7}\not \in B$ so B not group.

Why is the $B$ not group? Does the group homomorphism identity theorem work in modular arithmetic?

hhh
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  • What is $e$ supposed to be? – D_S Jan 11 '16 at 17:21
  • @D_S the euler's number, not the same as identity as used in the ProofWiki reference, updated. – hhh Jan 11 '16 at 17:37
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    Maybe I just can't see it, but is [tag:residue-calculus] actually appropriate? It looks like you're just talking about residue classes in the modular arithmetic sense. – pjs36 Jan 11 '16 at 18:02
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    @pjs36 thank you for the clarification, updated. – hhh Jan 11 '16 at 18:44

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If I'm not mistaken, $A$ is the subgroup of $\mathbb{Z}_{11}^{\ast}$ consisting of all quadratic residues, and $B$ is basically the same thing as $A$, except multiplication is now replaced by addition.

The function $A \rightarrow B$ which sends $x \mapsto e^x$ is not a group homomorphism, because $B$ is not a group! In fact, $B$ does not have an identity, and it is not closed under addition. So the theorem about 'the identity going to the identity' is meaningless here.

D_S
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  • Of course! Thank you, the initial assumption failed so I was unable to construct proper $\phi$. Is there any theorem to help to find the $\phi$ to test the theorem in practise? Does the preserving mean that there is the same number of identity elements in the group $G$ as in the group $H$ after the group homomorphism where the identity elements can be different? – hhh Jan 11 '16 at 17:27
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    Any group $G$ has one and only one identity element. – D_S Jan 11 '16 at 18:51