Let $G $ be an abelian group. In this section we shall use additive notation for $G$, so the group operation will be denoted by ' $+$ ' the identity by $0$, the inverse of $g$ by $-g$ and powers of $g$ by $2g, 3g, ....$

Let $2G$ be the subgroup of $G$ consisting of all elements of the form $g + g$ $(g \in G)$. If $G$ has a basis of $n$ elements, then $G/2G$ is a group of order $2^n$.

How the order of $G/2G$ is $2^n$? Plz show in an elementary way.

This an excerpt from "Algebraic number theory and Fermats last theorem by Ian Stewart, David Tall (3rd ed.).

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    $\begingroup$ $G$ is a vector space over the field of two elements. $\endgroup$ Aug 17, 2020 at 14:50

1 Answer 1


"... G has a basis of $n$ elements..." to me implies that $G$ is a free Abelian group of rank $n$, i.e. there is a basis made of $n$ linearly independent generators $g_1, g_2,\ldots,g_n\in G$; every element of $G$ can be uniquely represented as $m_1g_1+m_2g+\ldots+m_ng_n$ where $m_i\in\mathbb Z$ are "co-ordinates" in this basis.

(Recall we use the same terms in linear algebra, but the coefficients there come from a field, while here they come from the ring $\mathbb Z$ which is not a field. This makes calculations slightly more complicated.)

Now, note $2G$ is the subset of those elements of $G$ which have even co-ordinates, i.e. all $m_i$'s are even. That means that $G/2G$ is made up of "cosets" of the form $\lambda_1g_1+\ldots+\lambda_ng_n+2G$ where $\lambda_i\in\{0,1\}$. (Every $m_1g_1+\ldots+m_ng_n$ belongs to the coset $\lambda_1g_1+\ldots+\lambda_ng_n+2G$ where $\lambda_i =\text{ remainder in dividing }m_i\text{ by }2$). Thus, the number of those cosets (which is the order of $G/2G$) is $2^n$.

Hope this helps!

  • $\begingroup$ I agree that "basis" is not well-defined here, but perhaps it just means "generated by"? $\endgroup$ Aug 17, 2020 at 15:08
  • $\begingroup$ Not a helpful for a beginner....... lost from 2nd line...... $\endgroup$ Aug 17, 2020 at 15:09
  • $\begingroup$ @JasonDeVito There will be obvious counterexamples. $G=\mathbb Z_3$ is also generated by one element but $2G=G$ and $G/2G$ would be trivial. $\endgroup$
    – user700480
    Aug 17, 2020 at 15:10
  • $\begingroup$ @Andrew Terribly sorry. I don't have the book at hand... - How do they define "basis" of an Abelian group? - they use that term, they must've defined it somewhere. Happy to rewrite the proof in more elementary ways, but need first to understand "what I am allowed to use". $\endgroup$
    – user700480
    Aug 17, 2020 at 15:11
  • $\begingroup$ Oh, good point about the counterexamples. $\endgroup$ Aug 17, 2020 at 15:13

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