# How the order of $G/2G$ is $2^n$?

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.).

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

"... 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!

• I agree that "basis" is not well-defined here, but perhaps it just means "generated by"? Aug 17, 2020 at 15:08
• Not a helpful for a beginner....... lost from 2nd line...... Aug 17, 2020 at 15:09
• @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.
– user700480
Aug 17, 2020 at 15:10
• @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".
– user700480
Aug 17, 2020 at 15:11
• Oh, good point about the counterexamples. Aug 17, 2020 at 15:13