# The free group of rank $n$ mod its commutator subgroup is isomorphic to the free abelian group of rank $n$ [duplicate]

Show that: The free group of rank $n$ mod its commutator subgroup is isomorphic to the free abelian group of rank $n$.

I've tried to apply the first isomorphism theorem to this by defining the obvious homomorphism which takes generators to generators and trying to show that the commutator subgroup is equal to the kernel. It is trivial to see that the commutator subgroup is contained in the kernel, but I can't quite seem to show the other inclusion. A general element in the kernel would be a word $a_i a_j ... a_m$ where each generator occurs the same number of times as its inverse. How do I prove that this word is in the commutator subgroup?

Note: I have seen The quotient of a free group, of rank $n$, and its commutator subgroup is isomorphic to $\mathbb{Z}^n$ , but we have not covered the result that the answer there relies on (In any group $G$ and for any normal subgroup $H⊲G$ , we have that the quotient $G/H$ is abelian iff $G′≤H$), so I would appreciate an answer that does not rely on that.

• Can you show that you can rearrange the order of a word up to multiplication by commutators? – Cheerful Parsnip Nov 9 '14 at 21:16
• @GrumpyParsnip I'm not sure what you mean. Do you mean that I can rearrange the order by multiplying by commutators? How would that help me? – Johanna Nov 9 '14 at 21:21
• By rearranging, you can bring every term in a "normal form": $g_1^{n_1} g_2^{n_2} \cdots g_k^{n_k}$, where $g_1, g_2, \ldots, g_k$ are your generators, and $n_1, n_2, \ldots, n_k$ are integers. (In the case of infinitely many generators, be a bit more careful and define an ordered infinite product of elements all but finitely many of which are $1$.) But such an element can only lie in the kernel if $n_1 = n_2 = \ldots = n_k = 0$. – darij grinberg Nov 9 '14 at 21:25
• You have $a_1 a_2 \cdots a_k \equiv a_1 a_2 \cdots a_{i-1} a_{i+1} a_i a_{i+2} a_{i+3} \cdots a_k \pmod C$, where $C$ is the commutator subgroup. This way you can move any factor past any other as long as you only care about the $C$-coset. Now, move all $g_1$-factors and $g_1^{-1}$-factors to the very left, then move all $g_2$-factors and $g_2^{-1}$-factors to the right of it, etc. – darij grinberg Nov 9 '14 at 21:29
• Which part of the missing property did you not cover yet? Do you know that $G/G'$ is abelian? Do you see that for any homomorphism $G\to A$ with $A$ abelian, all elements of the form $xyx^{-1}y^{-1}$ must be in the kernel? – Hagen von Eitzen Nov 9 '14 at 21:55