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If a cyclic group has finite cardinality $k$ how do I prove that it is Isomorphic to the group $(\mathbb{Z}/k\mathbb{Z}, +)$.

I can show that the two groups have the same order but not sure how to prove bijectivity.

Shaun
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tomatoketchup
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  • Work out a mapping from one group to the other and prove that it is a group isomorphism. E.g. if cyclic group $G$ has generator $a$ of order $k$, pick a natural image for $a$ in $\mathbb Z/k\mathbb Z$ and extend this mapping to all of $G$. – hardmath May 07 '19 at 14:31
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    Are you sure that's your question? Bijectivity is trivial: two finite sets with the same number of elements is of course bijective! – balddraz May 07 '19 at 14:32
  • @ZeroXLR: If you know a map is onto in that situation, it is necessarily injective, and vice versa, but if you don't know either... – hardmath May 07 '19 at 14:33
  • consider mapping a generator – J. W. Tanner May 07 '19 at 14:34
  • @ZeroXLR For two sets, $A,B$, $|A|=|B|$ $|\iff$ $\exists$ $\psi : A \rightarrow B$ such that $imA=CodA$ and $\psi(a)=\psi(b) \implies$ $a=b$ $\forall a,b \in A$. Therefore if you define a map you need to show it is bijective, you can't assume that every map will be a bijection. –  May 07 '19 at 15:02
  • @topologicalmagician I was misled by the OP's wording. The OP said that the cyclic group had cardinality $k$ when he should have stuck to the phrase order. In the former case, obviously there is a bijection between that group and $\mathbb{Z}/k\mathbb{Z}$ which also has cardinality $k$. – balddraz May 07 '19 at 15:07

3 Answers3

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Let G be a cyclic group of finite order k. This means that $G=<g>$ for some $g\in G$. Define a map $\psi : \mathbb{Z}/k\mathbb Z \rightarrow G$ by $\psi(m)=g^{m}$. This map is an isomormphism. Note: to show bijectivity, it suffices to show injectivity or surjectivity; this is because the groups, as sets, are finite, and therefore the notions of surjectivity, bijectivity and injectivity are all equivalent when you're dealing with finite sets. Furthermore, you need to actually show that the map is a bijection rather than just say that because the sets are of equal length then that means the map mentioned above is a bijection. This is because two sets are of equal length if there is a bijection from one set to another, so you need to show that the map mentioned above is the map that exists (note: there can also be more than one, because the existential quantifier is not unique. THIS DOES NOT MEAN THAT EVERY MAP IS A BIJECTION).

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By definition of a cyclic group $G$, it's generated by a single element $g$. Since the size of your group is $k$, the order of $g$ must be $k$.

So, you'd like to identify this group with $\mathbb{Z}/k\mathbb{Z}$ with addition. A generator of $\mathbb{Z}/k\mathbb{Z}$ is $1$. Can you define a homomorphism from $G$ to $\mathbb{Z}/k\mathbb{Z}$ by sending one generator to another? Is it surjective? Is it injective?

To check the injectivity, think about where powers of $g$ go; if only $g^k$ goes to $0$, you're done. For surjectivity, you can write down an element going to any $m$ in $\mathbb{Z}/k\mathbb{Z}$.

J. W. Tanner
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probablystuck
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Hint $ $ Let $G = \langle g\rangle $. Group hom $\,h : \Bbb Z\to G,\ n\mapsto g^n$ is onto with kernel $\, k\Bbb Z\,$ by $\,g^n=1\!\iff\! k\mid n$ therefore by the First Isomorphism Theorem $\, G = {\rm Im}\,h\, \cong\, \Bbb Z/{\rm ker}\, h = \Bbb Z/k\Bbb Z$

Bill Dubuque
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