Quotient of $\Bbb Z^{\Bbb N}$ by a cyclic subgroup is isomorphic to itself

Question

Consider the group $G=\Bbb Z^{(\Bbb N)}$, the direct sum of countably many $\Bbb Z$'s. (Thus $\Bbb Z^{(\Bbb N)}$ is free abelian of rank $| \Bbb N|$.) Let $a$ be any nonzero element of $G$. Then $<a>$ is an infinite cylcic subgroup of $G$, isomorphic to $\Bbb Z$. I want to prove that $G/<a>$ is isomorphic to $G$, which seems obvious, but I don't know how to do. How do I have to do?

Answer 1

Say $a$ generates a maximal cyclic subgroup of $G=\Bbb{Z}^{\oplus \Bbb{N}}$ if for any other $b\in G$, $\langle a\rangle \subseteq \langle b\rangle$ implies $\langle b \rangle =\langle a \rangle$.

Then you can check that the following are equivalent

• $a$ generates a maximal cyclic subgroup of $G$,
• when $a=nb$, for $b\in G$, $n\in\newcommand\ZZ{\Bbb{Z}}\ZZ$, we have $n=\pm 1$,
• The greatest common divisor of the entries of $a$ is $1$.

Then $G/\langle a \rangle \cong G$ if and only if $a$ generates a maximal cyclic subgroup.

Proof.

First suppose $a$ generates a maximal cyclic subgroup of $G$. We'll prove that in fact $G$ has a free basis containing $a$, and thus that, up to an automorphism of $G$, we can take $a$ to be $(1,0,0,\ldots)$. To see that such a free basis exists, note that since $a$ has finitely many nonzero entries (which we may assume to be $1,\ldots,N-1$ without loss of generality), it suffices to find a free basis for $\ZZ^N$ containing $a$, since that can be extended to a free basis for $G$ by taking the standard basis vectors $e_n$ for $n>N$. We're using $N$ rather than $N-1$, since the extra coordinate, where we know $a_N=0$, allows us a little more flexibility and cleaner formulas.

Now since the greatest common divisor of the entries of $a$ is $1$, we can find integers $b_1,\ldots, b_{N-1}$ so that $$\sum_{i=1}^{N-1} a_ib_i = 1.$$ Then choose $f_i= e_i -b_ie_N$ for $1\le i \le N-1$.

Now observe that $$a - \sum_{i=1}^{N-1} a_if_i = \sum_{i=1}^{N-1} a_ib_i e_N = e_N.$$ Thus $\langle a,f_1,\ldots,f_{N-1}\rangle$ contains $e_N$, and thus also $e_i$ for all $1\le i\le N-1$, since $e_i=f_i+b_ie_N$. Hence $a,f_1,\ldots,f_{N-1}$ are $N$ elements of $\ZZ^N$ that generate $\ZZ^N$, and thus form a free basis, as desired.

Hence if $a$ generates a maximal cyclic subgroup of $G$, $G$ is the internal direct sum of $\langle a\rangle$ and a free subgroup isomorphic to $G$. Hence in this case, $G/\langle a \rangle \cong G$, as desired.

Now suppose $a$ doesn't generate a maximal cyclic subgroup. Then let $d$ be the greatest common divisor of the entries in $a$. Thus $a=db$, for some $b\in G$. Moreover, $b$ generates a maximal cyclic subgroup of $G$. But then $G\cong \langle b\rangle \oplus G'$, where $G'$ is a free subgroup isomorphic to $G$.

Hence $$G/\langle a\rangle \cong \left(\langle b\rangle /\langle a \rangle\right) \oplus G'\cong \ZZ/d\ZZ\oplus G.$$ Then since this group has nonzero torsion, it is not isomorphic to $G$. $\quad\blacksquare$