Question 1186053
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This a Abstract Algebra class
Let G=⟨a⟩ be a cyclic group of order10. Define φ:Z→G by φ(n)=a^n for all n∈Z. 
Assume that φ is a homomorphism. Find kerφ.
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<pre>
The group G is presented as  G = (a) in the post.

Also, from the context, G is a multiplicative group, regarding its operation.



It means that element "a" of the group G is the <U>generator</U> of the cyclic group G.



In turn, it means that  {{{a^10}}} = 1  (the unit element of the group G),  

while  the elements  {{{a^k}}}  at k = 1, 2, 3, . . . , 9 are all different and not equal to the unit element of the group.



Hence, kerφ is the set of all integers in Z that are multiples of 10:   { . . . , -20, -10, 0, 10, 20, . . . }.    <U>ANSWER</U>
</pre>

Solved and explained.



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We should not &nbsp;"assume" &nbsp;that  φ , &nbsp;defined in this way, &nbsp;is a homomorphism.


It &nbsp;REALLY &nbsp;IS &nbsp;the homomorphism.