Isomorphism to C_n with n prime

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SUMMARY

The discussion centers on proving that if the order of a group G is a prime number n, then G is isomorphic to the cyclic group of order n, denoted as C_n. The proof leverages Lagrange's theorem, which indicates that every element in G has an order of n when n is prime. The participants explore the implications of establishing a one-to-one mapping between G and C_n, concluding that mapping a single element from G to C_n determines the entire homomorphism due to the properties of cyclic groups.

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  • Understanding of group theory concepts, specifically cyclic groups.
  • Familiarity with Lagrange's theorem in group theory.
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Homework Statement



Prove taht if the order n of a group G is a prime number, then G must be isomorphic to the cyclic group fo order n, [tex]C_n[/tex].

The Attempt at a Solution



We have previously proven that a group can can be written as [tex]S = \{A,A^2,A^3,A^4...,A^n = E\}[/tex] where E is the identity and the group is of order n. We also have Lagrange that tells us, in this case, the order of every element in S is n if the order is prime.

So let's say we have the group [tex]G = \{A, A^2, A^3,..,A^g\}[/tex] where G is of order g which is prime and the cyclic group [tex]C_n = \{C, C^2, C^3,...,C^n\}[/tex] where n is again the prime order of the group. By this we know that [tex]A^m \ne E , m<g[/tex] and similarly [tex]C^m \ne E ,m<n[/tex].

Now it seems like you can make an absolutely arbitrary 1 to 1 mapping from [tex]G -> C_n[/tex], so is my best bet to try to prove that it's possible to make a non-1to1 mapping and show that it must not work?
 
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How much data about a homomorphism [tex]\varphi: G \to C_n[/tex] do you need to completely determine [tex]\varphi[/tex]? It isn't much.
 
I suppose you just need to map one element and then by construction, the rest is figured out. That is, if I pick some [tex]A_1 \epsilon G[/tex] and map it to some [tex]C_1 \epsilon C_n[/tex], then I should be able to say [tex]A_1^2 = A_2 = C_1^2=C_2[/tex] which gives me a unique mapping I would think.
 

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