TITLE: Can all squares in a free group be made from squares in the free monoid?
QUESTION [13 upvotes]: Here's a question I thought of, that I don't know the answer to.
Let $F_2$ be the free group on $\{a,b\}$, and $F_2^+$ be the subset where all the exponents are positive. For a set $S$, let $S^2=\{g^2:g\in S\}$.
Is $\langle (F_2^+)^2\rangle$ (the group generated by $(F_2^+)^2$) the same as $\langle (F_2)^2\rangle$?
For example, can $(a^{-1}b)^2$ be written as a product of squares, each with all-positive or all-negative exponents?
I've thought a bit about Cayley diagrams, but I'm not sure how to make progress.

REPLY [10 votes]: A less algorithmic approach, but a little simpler.
You only really need to prove that $K=\langle (F_2^+)^2\rangle$ is a normal subgroup of $F_2.$
That’s because,  $H=\langle F_2^2\rangle$ is a normal subgroup of $F_2.$ We have $F_2/H\equiv C_2^2.$
$K$ is a subgroup of $H$ which contains $a^2, b^2,$ and $(ab)^2.$ But if $K$ is normal, this means we have $F_2/K=C_2^2,$ since $aK,bK, abK$ all have order $2$ in $F_2/K.$ Since $K\subset H,$ $K=H.$

Now, it is enough to prove if $w\in F_2^+,$ then $$aw^2a^{-1},bw^2b^{-1}\in K.$$
But we can write $$aw^2a^{-1}=(w^2)^{-1}(w^2a)^2(a^2)^{-1}\in K$$ since $w,a\in F_2^+,$ we also have $w^2a\in F_2^+.$ And similarly for $bw^2b^{-1}\in K.$
From this we see that:
$$a^{-1}w^2a=(a^2)^{-1}(aw^2a^{-1})a^2\in K,$$ and similarly for $b^{-1}w^2b.$
But any subgroup of $F_2$ with $aKa^{-1}=K$ and $bKb^{-1}=K$ is a normal subgroup of $F_2.$

The formula for $aw^2a^{-1}$ is an extension of a comment to the other answer, where it was given for $w=b.$

We can make this algorithmic. An element is in $H$ if it can be written as a word with an even number of $a,a^{-1}$ terms and an even number of $b,b^{-1}$ terms.
If the word starts with $aa,bb,a^{-1}a^{-1},$ or $b^{-1}b^{-1}$ then we can shorten the word.
If the word starts with $aba,$ say $w=abaw_1$ then we can write it as $$w=(abab)(b^{-1}w_1)$$ where $b^{-1}w_1\in H$ has a shorter length.
So given any four letter word in $H,$ we need an expression as a product of elements of $(S^+)^2.$ You can restrict to such words that end in $a$ or $b.$ And, by the automorphism of $F_2$ swapping $a,b,$ we can do only the words ending in $b.$
We finally are down to the twelve cases:
$$
\begin{align}
abab&=(ab)^2
\\aba^{-1}b&=(ab)^2 a^2(ba^2)^{-2}b^2
\\ab^{-1}ab&=a^2 (ba)^{-2} a^{-2}(a^2b)^{2}
\\ab^{-1}a^{-1}b&=a^2(ba)^{-2}b^2
\\a^{-1}bab&=a^{-2}(ab)^2
\\a^{-1}ba^{-1}b&=a^{-2} (ab)^2 a^2(ba^2)^{-2}b^2
\\a^{-1}b^{-1}ab&=(ba)^{-2} a^{-2}(a^2b)^{2}
\\a^{-1}b^{-1}a^{-1}b&=(ba)^{-2} b^2\\
baab&= a^{-2}(a^2b)^{2}\\
ba^{-1}a^{-1}b&=b^2 a^2(ba^2)^{-2}b^2\\
b^{-1}aab&=b^{-2}a^{-2}(a^2b)^{2}\\
b^{-1}a^{-1}a^{-1}b&=a^2(ba^2)^{-2}b^2
\end{align} 
$$
This means that $H$ is generated by $$a^2,b^2,(ab)^2,(ba)^2, (a^2b)^2, (ab^2)^2, (b^2a)^2,(ba^2)^2\in (F_2^+)^2$$
We can reduce this somewhat because:
$$(b^2a)^2=bb(abba)=b^{2}(ab^2)^2b^{-2}\\
(ba^2)^2=(baab)aa=a^{-2}(a^2b)^2a^2\\ (a^2b)^2=a^2 (ba)^2 b^{2}(ab^2)^{-2} (ab)^2 $$
So we are down to five generators:
$$a^2,b^2,(ab)^2,(ba)^2, (ab^2)^2\in (F_2^+)^2$$
We can prove this set of generators generates all of $H$ by proving formula for $aga^{-1},bgb^{-1}$ when $g$ is in the generator set. We’ll just show for $a.$
$$
\begin{align} 
a(a^2)a^{-1}&=a^2\\
a(b^2)a^{-1}&=(ab^2)^2b^{-2}a^{-2}\\
a(ab)^2a^{-1}&=a^2(ba)^2a^{-2}\\
a(ba)^2a^{-1}&=(ba)^2\\
a(ab^2)^2a^{-1}&=a^2b^2(ab^2)^2b^{-2}a^{-2}
\end{align} $$
All the cases for $b$ are dual, except:
$$
\begin{align} 
b(ab^2)^2b^{-1}&=(ba)^2a^{-2}(ab)^2b^{-2}
\end{align}$$
Generators reduced to $5$ by the accepted answer to this question.