Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that we have a free abelian group $F$. How can it be proved that $F$ has a subgroup of index $n$ which $n≥1$?

Honestly, according to the Theorems, I just know that if we take $X$ as a base for $F$, then $$ F= \bigoplus_{\alpha \in X} \mathbb Z_\alpha \ $$ in which for all $ \alpha \in X$; $\mathbb Z_\alpha \ $ is a copy of $ \mathbb Z $. What that subgroup could be? Thanks.

share|cite|improve this question
Looks like you are referring to free abelian groups? – Grumpy Parsnip May 16 '12 at 15:43
What about taking the subgroup of $\mathbb{Z}$ generated by $n$ in the first copy, and $\mathbb{Z}$ in the other copies. Would that work? – M Turgeon May 16 '12 at 15:43
Whether he means abelian or non-abelian free groups, answering it for free abelian groups is enough, since any non-trivial free group has a non-trivial free abelian group as a quotient. – Thomas Andrews May 16 '12 at 15:58
@BabakSorouh He is saying take some $\alpha_0\in X$ and define $H_\alpha$ to be $\mathbb Z_\alpha$ for $\alpha\neq \alpha_0$ and take $H_{\alpha_0}=n\mathbb Z_{\alpha_0}$. Then define $$H = \bigoplus_{\alpha\in X} H_\alpha$$ – Thomas Andrews May 16 '12 at 16:03
@Babak: Then say "free abelian". If you say "free group", you mean the absolutely free group. – Arturo Magidin May 16 '12 at 16:20
up vote 2 down vote accepted

Theorem. Let $\mathfrak{V}$ be a variety of groups, and let $X$ be a nonempty set. The free $\mathfrak{V}$-group on $X$, $F_{\mathfrak{V}}(X)$ has a subgroup of index $n$ if and only if there is some $|X|$-generated group in $\mathfrak{V}$ with a subgroup of index $n$.

Proof. If $F_{\mathfrak{V}}(X)$ has a subgroup of index $n$, then it witnesses the existence of such a group. Conversely, let $G\in\mathfrak{V}$ be a group, with $\{g_x\}_{x\in X}$ a generating set of $G$, and suppose that $H$ is a subgroup of $G$ of index $n$. The map $f\colon X\to \{g_x\}_{x\in X}$ given by $f(x)=g_x$ induces, by the universal property, a surjective homomorphism $\mathfrak{f}\colon F_{\mathfrak{V}}(X)\to G$. By the isomorphism theorems, $H$ corresponds to a subgroup $\mathcal{H}$ of $F_{\mathfrak{V}}(X)$ that contains $\mathrm{ker}(\mathfrak{f})$, and hence $[F_{\mathfrak{V}}(X):\mathcal{H}] = [G:H] = n$, as claimed. $\Box$

Corollary. If $X\neq\varnothing$, then the free abelian group on $X$ has a subgroup of index $n$ for every positive integer $n$.

Proof. Let $C_n$ be the cyclic group of order $n$. This is $|X|$ generated, and contains a subgroup of index $n$ (namely, $\{1\}$). $\Box$

share|cite|improve this answer
Thank you dear Prof. for the answer. May I ask you " if we have this subgroup, then $F$ is an extention of $H$ by $n\mathbb Z_{\alpha_{0}}$" as comments above? Thanks again for the time. – Babak S. May 16 '12 at 16:49
@Babak: There is no such thing as "this subgroup". There are infinitely many different subgroups of index $n$. Any subgroup of a free abelian group is necessarily free abelian, and a quotient of $F$ by a subgroup of index $n$ is necessarily of order $n$, and so in particular it cannot be torsion free (as it would be if it were $n\mathbb{Z}_{\alpha_0}$. You simply have completely misunderstood the comments above. It seems you just don't understand what is going on at all. – Arturo Magidin May 16 '12 at 17:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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