# $(G:H) < \infty$ where $G$ is an infinite cyclic group

I'm learning Abstract Algebra, specifically cyclic groups, and need help with the following problem:

Let $G$ be an infinite cyclic group and $\{1_{G}\} \neq H \leq G$. Show that $(G:H) < \infty$.

Since I'm new to this subject, I first tried to rephrase the problem with my own words. If I understand it correctly, I need to show that for a non-trivial subgroup $H$ of an infinite cyclic group $G$, the index of $H$ in $G$ is finite.

The only relation I could make with the theory is that since $G$ is an infinite cyclic group, then the subgroup $H$ of $G$ is also cyclic. Also, I know that the index of $H$ in $G$ is closely related to Lagrange's Theorem but I don't know if it is useful here.

• Since $H$ is not trivial, there is an element $1 \neq y \in H$. We have that $y = x^k$ for some $k \in \mathbb{Z}_0$ where $G$ has generator $x$. Consider the minimal positive exponent of elements in $H$, say $n$ then $H$ is generated by $x^n$ and the index is $n$, hence finite. – Student Feb 12 '17 at 11:14
• @Student I'm having difficulties understanding the second part of your argument. What do you mean by: "Consider the minimal positive exponent of elements in $H$"? Also, the definition I'm familiar for the index of $H$ in $G$ is the number of left (or right) cosets of $H$. You were able to determine the index of $H$ without counting the numbers of cosets of $H$ and I didn't get that. – user347616 Feb 12 '17 at 11:52
• All elements in $H$ are of the form $x^k$ for some integer $k$. Consider the set of strictly positive exponents of elements in $H$. This is a subset of the natural numbers, which is bounded bellow by $1$, hence there is a minimal element, say $n$. You can prove (using the division algorithm) that $H$ is generated by $x^n$. It is now quite easy to see that the possible cosets are $H, x \ast H, \ldots x^{n-1} \ast H$ where $\ast$ is the group operation. – Student Feb 12 '17 at 11:58

Let g be a generator of $G$ then for some $k$ we have $g^k \in H$ so there are no more than $k$ cosets of $H$, i.e. $Hg^k = H$. .
As you pointed out, $H$ will also be a cyclic group, say $\langle y \rangle = H$ (letting for instance $G= \langle x \rangle$). Then $y = x^n$ for some $n$.
So $x^k = x^m (mod H)$ if and only if $x^{k-m} \in H$ if and only if $n | k-m$ if and only if $k = m [n]$. How many equivalence classes mod $H$ are there ?
($n \neq 0$ because $H\neq \{1_G\}$)