Describe the class of groups that satisfy $(x^2y^2)^2 \approx 1$

Question

I have trouble with the following assignment:

Let $A$ denote the class of all Abelian groups satisfying the identity $x^2 \approx 1$. Show that the class $$\{G \mid \exists N: N \trianglelefteq G \land N \in A \land G/N \in A\}$$ is precisely the class (variety) of all groups satisfying the identity $(x^2y^2)^2 \approx 1$

I have no trouble showing that any group from the first class satisfies the identity in the second class, but I can not show the converse.

Any time I come up with a subgroup it either fails to be normal (or I just can’t prove it), or it is normal subgroup and is element of $A$, but the quotient group fails to be in $A$ (or I can not prove it to be true).

My best attempts:

$$N = \{ x \in G \mid x^2 = 1 \land \forall y \in G: xy=yx\}$$ One can easily verify that this is a normal subgroup of $G$ that is in $A$, but I don’t think that $G/N \in A$ because $\forall y \in G/N \,\exists g \in G: y = Ng$. Then $y^2 = 1$ iff $(Ng)^2 = N$ iff $N(g^2)=N$ iff $g^2 \in N$ (I don’t think this is true in general).

So from this maybe we can force $N$ to contain $\{x^2\mid x \in G\}$ by taking something like $N = \langle\{x^2\mid x \in G\}\rangle$, but I dont see how this would be normal or satisfy the identity $x^2 \approx 1$.

Thank you for suggestion!

Answer 1

Let $G$ be a group satisfying the identity $(x^2y^2)^2\approx 1$. Let $N$ be the subgroup generated by all squares. As this is a verbal subgroup, it is normal (in fact, fully invariant). If you don't know this, note that the conjugate of a square is a square, so for $X=\{x^2\mid x\in G\}$, we have $gXg^{-1}\subseteq X$ for all $g\in G$, so $\langle X\rangle$ is normal.

Then $G/N$ has exponent $2$ and is therefore abelian. (None of this uses that $G$ satisfies the identity...)

Note that the elements of $G$ all have exponent $4$: if $x\in G$, then taking $y=e$ gives $(x^2)^2=1$.

And since $G$ satisfies the identity, the product of any two squares has exponent $2$. In particular, squares commute: since $x^2y^2x^2y^2=1$, and $x^{-2}=x^2$ for all $x$, we have $x^2y^2=y^{-2}x^{-2}=y^2x^2$.

Thus, $N$ is abelian (it is generated by pairwise commuting elements), generated by elements of order $2$, hence satisfies the identity $x^2=1$.

Answer 2

Notice that if $G$ has the property that $N$ and $G/N$ are boolean (=satisfy $x^2 \approx 1$), then $G$ actually satisfies the identities $(x_1^2 \dotsc x_n^2)^2 \approx 1$ for every $n \geq 0$. For this, I can at least prove the converse. This here is therefore just a partial answer (for now, maybe I can complete it later).

If $(x_1^2 \dotsc x_n^2)^2 = 1$ holds for all $x_1,\dotsc,x_n \in G$, let $N := \langle x^2 : x \in G \rangle$. Explicitly, we have $N = \{x_1^2 \dotsc x_n^2 : x_1,\dotsc,x_n \in G \}$. This is clearly a characteristic subgroup, hence normal, and the assumption precisely means that $N$ is a boolean group. Also, since $N$ contains all squares, $G/N$ is boolean as well.

Therefore, our task is to derive from $(x^2 y^2)^2 \approx 1$ the general identity $(x_1^2 \dotsc x_n^2)^2 \approx 1$. EDIT. This is done in Arturo's answer.