6

If $F$ is a free group on a finite set $S$, then the squares in $F$ generate a normal subgroup $N$ and $F/N$ is elementary abelian $2$-group of order $2^{|S|}$.

Let $F$ be free group on infinite set $S$, and $N$ the normal subgroup generated by squares in $F$.

Q. $F/N$ is infinite direct sum or product of $|S|$ many copies of $\mathbb{Z}_2$?

p Groups
  • 10,228
  • I'm not sure about the first statement itself, namely that the squares in $F$ generate a normal subgroup $N$. Or do you mean $N$ is the smallest normal subgroup containing the group generated by squares? – Seven Apr 22 '16 at 05:35
  • 2
    Since $F$ is composed of finite words, I would bet on the direct sum. – Quang Hoang Apr 22 '16 at 05:41
  • @joriki Oh, got it. I was assuming that $N$ is generated by the squares of the generators. – Seven Apr 22 '16 at 05:41
  • 1
    @Seven: They do generate a normal subgroup, as e.g. $g^{-1}aabbg = \left(g^{-1}ag\right)\left(g^{-1}ag\right) \left(g^{-1}bg\right) \left(g^{-1}bg\right)$. – joriki Apr 22 '16 at 05:41
  • @Steven: generate mens take all possible products of squares in $F$; we certainly get a subgroup. – p Groups Apr 22 '16 at 06:47

1 Answers1

3

You could conclude easily by saying that since the words are finite, then you must be in the direct sum and not the direct product as Quang Hoang points out.

But I think the most adequate point of view in general is that of universal properties (it's the most helpful in general cases) : $F$ has the universal property the any function $f:S\to G$ to a group $G$ extends uniquely to a morphism $\tilde{f}: F\to G$. Then the quotient $F/N$ has the property that any group morphism $\tilde{f}: F\to G$ such that $\tilde{f}(a)^2 = 1$ for all $a\in S$ factors uniquely to a morphism $\overline{f}:F/N\to G$.

So $F/N$ has the property that any function $f:S\to G$ such that $f(a)^2= 1$ for all $a\in S$ extends uniquely to a group morphism $\overline{f}: F/N\to G$.

This is exactly the universal property of $\bigoplus_S \mathbb{Z}/2\mathbb{Z}$, since such a function $f:S\to G$ is the same as giving a morphism $f_a: \mathbb{Z}/2\mathbb{Z}\to G$ for all $a\in S$.

So $F/N$ and $\bigoplus_S \mathbb{Z}/2\mathbb{Z}$ have the same universal property : they are canonically equal.

Captain Lama
  • 25,743