# A type of isomorphism on $\mathbb{Z}^r$

Let $(x_1,\dots,x_r)$ be a non-zero element of $\mathbb{Z}^r$, and let $h$ be the highest common factor of $x_1, \dots, x_r$. Show that there is an isomorphism $\mathbb{Z}^r \to \mathbb{Z}^r$ taking $(x_1,\dots,x_r)$ to $(1, 0, 0,\dots,0)$ if and only if $h=1$.

One direction isn't too bad; if $\phi$ is such an isomorphism then $$\phi\big((x_1,\dots,x_r)\big) = (1,0,\dots,0)$$ $$\phi\big(h(\tfrac{x_1}{h},\dots,\tfrac{x_r}{h})\big) = (1,0,\dots,0)$$ $$h \phi\big((\tfrac{x_1}{h},\dots,\tfrac{x_r}{h})\big) = (1,0,\dots,0).$$ But then $h|1$, so we must have $h=1$. I really have no idea how to go about proving the other direction... so any help would be appreciated.

-
Homomorphisms can only increase the (absolute value of the) gcd, so isomorphisms must preserve it. – Qiaochu Yuan May 29 '11 at 20:19

Just go through the extended Euclidean algorithm for expressing $h$ as an integer linear combinantion of $x_1,\ldots,x_r$. Try the case $r=2$ and see how $(x_1,x_2)$ is reduced to $(h,0)$.

-
Okay.. treat me a little like an idiot for the moment because I am still not sure what $\phi$ does. So, assuming $\phi$ takes $(x_1,x_2)$ to $(h,0)$, where $h=ax_1+ bx_2$ for some integers $a,b$, what does it do to other elements? It is actually possible to express $\phi$ algebraically, say? – Sputnik May 30 '11 at 13:17
@Fadah, $\phi(1,0)=(a,-x_2)$, $\phi(0,1)=(b,x_1)$, for instance. Note that the determinant is 1, which means $\phi$ is invertible. – lhf May 30 '11 at 22:03

If there is an isomorphism $h$ mapping $(x_1,\dots,x_n)$ to $(1,0,\dots,0)$, its inverse $h^{-1}$ maps $(1,0,\dots,0)$ to $(x_1,\dots,x_n)$, and the matrix $A$ of $h^{-1}$ with respect to the canonical basis of $\mathbb Z^n$ then has $(x_1,\dots,x_n)$ as its first column. Since $A$ must have determinant $1$ or $-1$, the $\gcd$ of each of its columns is $1$.

-

Let $H$ be the cyclic subgroup of $\mathbb{Z}^r$ generated by $(x_1,\ldots,x_r)$. If $h=1$, then $t(y_1,\ldots,y_r) = (x_1,\ldots,x_r)$ implies $t = \pm{1}$, and so $\mathbb{Z}^r/H$ is torsion-free and hence free abelian. So $H$ has a complement in $\mathbb{Z}^r$, which must be isomorphic to $\mathbb{Z}^{r-1}$, and the result follows easily.

-