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I am trying to show that the quotient $$\frac{\langle a_1,\ldots,a_n\rangle}{\langle a_2-a_1,a_3-a_2,\ldots,a_{n}-a_{n-1},a_1-a_n\rangle}\cong \mathbb{Z}.$$

Is the following argument correct?

Since $(a_2-a_1)+\ldots+(a_n-a_{n-1})=-(a_1-a_n)$, the generator $a_1-a_n$ in $\langle a_2-a_1,a_3-a_2,\ldots,a_{n}-a_{n-1},a_1-a_n\rangle$ is redundant. I can show that none of the other generators $a_2-a_1,\ldots,a_{n}-a_{n-1}$ are redundant by induction, since if the first $n-1$ of them are not redundant, then the $n$th generator introduces $a_n$ which is not in any of the other generators. Therefore, $$\frac{\langle a_1,\ldots,a_n\rangle}{\langle a_2-a_1,a_3-a_2,\ldots,a_{n}-a_{n-1},a_1-a_n\rangle}\cong \frac{\mathbb{Z}^n}{\mathbb{Z}^{n-1}}\cong\mathbb{Z}.$$

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  • $\begingroup$ I'd find mutually inverse maps between both sides. $\endgroup$ – Lord Shark the Unknown Oct 12 at 10:48
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    $\begingroup$ Your argument doesn't work, as the isomorphism class of the kernel does not uniquely define the image. For example, replacing each $a_i$ with a $2a_i$ preserves the isomorphism class of the kernel but clearly alters the image. $\endgroup$ – user1729 Oct 12 at 16:17
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Hint: In the free group $G$ generated by $a_1,\ldots,a_n$, let $I$ be the normal subgroup generated by $a_i-a_1$, $1\le i\le n$. Then in quotient group $G/I$, $a_i \in a_1+I$, $1\le i\le n$, and so each element of $G/I$ has the form $ka_1+I$ for some integer $k$.

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