Exercise 2.3.17 (Presentation of $Z_n$)

Proof. Let $Z_n=⟨x⟩$, where $|x|=n$. We show that

Let $S=\{a\}$, and consider the free group $F(S)=F(a)$ generated by $a$. Then $F(a)\simeq$. Moreover since $F(a)$ is abelian, every subgroup is normal, so the normal subgroup closure of $R=\{a^n\}$ is $N=⟨a^n⟩$ (every subgroup of $F(a)$ which contains $a^n$ contains $⟨a^n⟩$).

By definition of a presentation,

Let $y=\bar{a}(\mathit{mod}⟨a^n⟩)$. Then $|y|=n$: indeed, $y^n=1$ and if $y^k=1$, then $a^k\in ⟨a^n⟩$, thus $a^k={(a^n)}^l$ for some integer $l$, so $a^{k-\mathit{nl}}=1$. Since $F(a)$ is a free group, $a$ does not satisfy any nontrivial relation, so $k-\mathit{nl}=0$, therefore $k=\mathit{nl}$ and $n\mid k$. This shows that $|y|=n$.

Then $F(a)∕⟨a^n⟩=\{1,y,y^2,\dots ,y^{n-1}\}$ is the cyclic group $⟨y⟩$, which is isomorphic to $Z_n$, by the isomorphism $Z_n\to F(a)∕N$ which maps $x$ on $y=\bar{a}$. So

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