# Rank of free group

Let $G$ be a free group with a basis $S$.

Let $G'$ be the commutator subgroup of $G$.

Define $S'=\{gG'\in G/G': g\in S\}$

Then, $S'$ is a basis for the free abelian group $G/G'$.

Following the above argument, How do I prove that $|S|=|S'|$?

• If not, then you would have $g,h \in S$ with $gh^{-1} \in G'$, which is impossible because elements in $G'$ have zero exponent sum for all elements of $S$. – Derek Holt Apr 4 '15 at 7:38
• @DerekHolt Doesn't $gh^{-1}$ have zero exponent sum as well? – Theo Apr 4 '15 at 7:40
• @Theo No, I meant the exponent sum of each individual generator. The exponent sums of $g$ and $h$ in $gh^{-1}$ are $1$ and $-1$. But for commutators, and hence for all elements of $G'$, the exponent sum of each generator is $0$. – Derek Holt Apr 4 '15 at 7:52
• Right, right, apologies... – Theo Apr 4 '15 at 7:56

## 1 Answer

Free group. Let $G$ be the free group on $S$. It consists of words ("strings") over the alphabet $S$. For instance $S = \{a,b\}$. Then if $x \in$ the commutator subgroup $G'$, then $xy = yx \ \forall y \in G$. But take $y = aaa$, $x = b$. Then $xy \neq yx$. Thus if $|S| \gt 1$, $G' = \{1\}$, and if $|S| = 1, \ G' = G$. Is that correct? Then $G/G' \approx G$ in the first case and $G/ G' \approx \{1\}$ in the second case.