Can all squares in a free group be made from squares in the free monoid?

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.

Let $F_2^-$ denote the subset where all the exponents are negative.

Lemma: $\langle(F_2^+)^2\rangle=\langle(F_2^-)^2\rangle$

Let $x^2$ be a generator in $\langle(F_2^+)^2\rangle$ where $x\in F_2^+$. Then the inverse $(x^2)^{-1}=(x^{-1})^2$ is a word in $\langle(F_2^-)^2\rangle$. $\langle(F_2^-)^2\rangle$ is a group so it contains the inverse of $(x^2)^{-1}$ which is $x^2$. Thus, $\langle(F_2^+)^2\rangle\subseteq\langle(F_2^-)^2\rangle$. The proof of the reverse inclusion is symmetrical.

Main Proof:

We clearly have $\langle(F_2^+)^2\rangle\subseteq\langle(F_2)^2\rangle$. To prove the reverse inclusion, we show that for any generator $w$ in $\langle(F_2)^2\rangle$ we can algorithmically multiply $w$ by elements of $\langle(F_2^+)^2\rangle$ to obtain the identity element. Since $\langle(F_2^+)^2\rangle$ is a group, this would show we can generate $w$ using elements from $\langle(F_2^+)^2\rangle$.

Let $W$ be a generator of $\langle(F_2)^2\rangle$. We say that a word $W^*$ in $F_2$ has positive parity if $W^*\in F_2^+$, has negative parity if $W^*\in F_2^-$, and has no parity if otherwise.

Step $1$: Transform $W$ into a word that has parity

At each step if $W$ still does not have parity, rewrite it in the form $W=XyZ$ where $X$ is the max-length sequence of characters at the beginning of $W$ such that $X$ is a word with parity. $y$ is a character (either $a, b, a^{-1}, b^{-1}$) that has different parity to $X$. $Z$ is the sequence of remaining characters in $W$. Left multiply $W$ by $(X^{-1})^2$ to obtain $X^{-1}yZ$ and notice that $X^{-1}y$ is a word that has parity and is longer than $X$. After enough transformations, we will have transformed the original word $W$ into a new word $W_2$ which has parity.

Step $2$: Remove all nested squares inside $W_2$

We say that $W_2$ has nested squares if we can write it of the form $xc^2y$ where $x,y\in F_2$ and $c$ is a character in $F_2$ (either $a, b, a^{-1}, b^{-1}$). Left multiply by $t=x^2c^2[(xc^2)^{-1}]^2$ to obtain $xy$ (note that $xy$ has parity). Note that as a result of Step $1$ we know that $x$ and $c$ have the same parity so $t$ is an element of $\langle(F_2^+)^2\rangle$. Eventually we obtain a transformed word $W_3$ that is free of nested squares and has parity.

Final Step: Removing $(ab)^2$

Note that every character $c$ appearing in the original word $W$ appears an even number of times since $W$ is the square of a word in $F_2$. After Step $1$ we obtained a word $W_2$ with parity. W.L.O.G. we can assume $W_2$ has positive parity since if had negative parity, we could just left multiply by $(W_2^{-1})^2$ to obtain $W_2^{-1}$. The character $a$ appears an even number of times in $W_2$, and the character $b$ appears an even number of times in $W_2$. Step $2$ only deletes characters two at a time, so the same thing is true for $W_3$.

Let $n_a$ and $n_b$ denote the number of times respectively that the characters $a$ and $b$ appear in $W_3$. If $n_a$ and $n_b$ are both $0$ then $W_3=1$ and we are done. If one of them is $0$ and the other is nonzero, W.L.O.G. let's say $n_a\neq 0$ and $n_b=0$. Then since $n_a$ is even there must be a nested square in $W_3$, contradicting Step $2$. So now we're left with the case that $n_a\neq 0$ and $n_b\neq 0$. W.L.O.G. let's say the first character of $W_3$ is $a$. Then because $n_a$/$n_b$ are even and $W_3$ is square-free, we must be able to write $W_3$ as $W_3=(ab)^n$ where $n$ is an even number. Left multiply by $(ab)^{-n}$ to obtain the identity.

Let me know if there's any part of the proof you want me to clarify. As an example, here's the way you were trying to find to express $(a^{-1}b)^2$.

$$a^{-1}ba^{-1}b$$ $$aba^{-1}b$$ $$b^{-1}a^{-1}a^{-1}b$$ $$aabb$$ So $$(a^{-1}b)^2=[a^{-1}]^2(ab)^2[(aab)^{-1}]^2a^2b^2$$

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.