If G is cyclic, and G is isomorphic to G', then G' is cycli

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SUMMARY

If a group G is cyclic and isomorphic to another group G', then G' is also cyclic. This conclusion is derived from the definition of cyclic groups, where for every element b in G, there exists an element a such that a^n = b for some integer n. The isomorphism φ from G to G' preserves the cyclic structure, confirming that G' inherits the cyclic property. Additionally, it is noted that surjectivity alone suffices for this proof, and that all subgroups of cyclic groups are also cyclic.

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Mr Davis 97
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Homework Statement


Title

Homework Equations

The Attempt at a Solution


This would seem to be very easy problem, since it's intuitively obvious that if two groups are isomorphic, and one is cyclic, then the other is cyclic too. However, I can't seem to formalize it with math.

Here is an idea. We can define that a group is cyclic by saying that for all b in G, there exists an a such that ##a^n = b## for some integer n. Now if G is cyclic, and we have an isomorphism ##\phi## from G to G', then it is true that ##\phi (a) ^n = \phi (b) = b'##, which means that G' is also cyclic. Does this sketch of a proof the right idea?
 
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Mr Davis 97 said:

Homework Statement


Title

Homework Equations

The Attempt at a Solution


This would seem to be very easy problem, since it's intuitively obvious that if two groups are isomorphic, and one is cyclic, then the other is cyclic too. However, I can't seem to formalize it with math.

Here is an idea. We can define that a group is cyclic by saying that for all b in G, there exists an a such that ##a^n = b## for some integer n. Now if G is cyclic, and we have an isomorphism ##\phi## from G to G', then it is true that ##\phi (a) ^n = \phi (b) = b'##, which means that G' is also cyclic. Does this sketch of a proof the right idea?
Yes it is. You can leave out the ##b's## as you don't need them, the generator ##a## is sufficient, i.e. simply write ##a^n##. It doesn't need to be named. And you may forget the isomorphism, surjectivity is sufficient. And subgroups of cyclic groups are also cyclic.
 

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