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, $$\frac{\langle a_1,\ldots,a_n\rangle}{\langle a_2-a_1,a_3-a_2,\ldots,a_{n}-a_{n-1},a_1-a_n\rangle}\cong \frac{\mathbb{Z}^n}{\mathbb{Z}^{n-1}}\cong\mathbb{Z}.$$

  • $\begingroup$ I'd find mutually inverse maps between both sides. $\endgroup$ – Lord Shark the Unknown Oct 12 at 10:48
  • 1
    $\begingroup$ 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

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 $k$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.