# Quotient of free groups $\langle a_1,\ldots, a_n\rangle /\langle a_2-a_1,a_3-a_2,\ldots,a_{n}-a_{n-1},a_1-a_n\rangle$

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

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