Cfrgtkky

Can someone give an example of a finite (ideally nonabelian) group $G$ and two surjective homomorphisms
$phi_1,phi_2 : F_2 rightarrow G$ (where $F_2$ is the free group on the generators $x,y$), such that $phi_1(x^{-1}y^{-1}xy)$ and $phi_2(x^{-1}y^{-1}xy)$ have different orders?

Is this possible?

Are there conditions on $G$ that make this impossible? (In other words, as suggested by Mariano Suárez-Alvarez, are there conditions on $G$, still assumed to be a 2-generated group such that the commutator $[a,b]$ always has the same order for any pair of generators $a,b$?)

thanks

Consider the elements $a=(12345678)$, $b=(12)$ and $c=(12)(34)$.

The sets ${a,b}$ and ${a,c}$ both generate $S_8$. Compute the orders of the corresponding commutators.

A somewhat trivial general sufficient condition is that $G'$ is cyclic of prime order $p$. Then the commutator of any pair of generators is an element of order $p$. (The two non-abelian groups of order $p^{3}$ provide examples here.)

This is not necessary, however. In the group
$$
G = langle x, y : x^{p^{3}} = y^{p^{2}} = 1, [x, y] = x^{p} rangle,
$$
where $p$ is a prime (perhaps an odd one), if $G = langle a, b rangle$, then $[a, b]$ has order $p^{2}$.

The reason is that in a (non-abelian, say) group $H = langle a, b rangle$ we have that $H'$ is the normal closure of $langle [a, b] rangle$. In this particular group $G$, we have that $G' = langle a^{p} rangle$ is cyclic of order $p^{2}$, with all of its subgroups ${1}, langle a^{p^{2}} rangle, langle a^{p} rangle$ characteristic in $G$. So if for two generators $a, b$ we have that $[a,b]$ has order less than $p^{2}$, then $[a, b] le langle a^{p^{2}} rangle$, and the normal closure of $langle [a, b] rangle$ is definitely not $G'$.

So a more general sufficient condition is that the derived subgroup of the two-generator group is cyclic, because all the subgroups of a cyclic group are characteristic. (So we get more examples here, for instance the dihedral groups.)

Once more, this is not necessary, as the example of $A_{4}$ shows. This example suggests that another sufficient condition is that $G'$ is elementary abelian, so that all of its non-trivial elements have the same prime order. (Examples here are provided by the affine group of a finite line, that is, the semidirect products $F rtimes F^{star}$, where $F$ is the additive groups of a finite field, and $F^{star}$ its multiplicative group, acting on $F$ by multiplication. When $F = operatorname{GF}(4)$, we get $A_{4}$.)

Addendum I extract from the comments below an observation. Suppose $G'$ is a $p$-group for some prime $p$. Then $G$ satisfies the condition provided the following holds:

if the exponent of $G'$ is $p^{e}$, then $Omega_{e-1}(G')$ is a proper subgroup of $G'$.

Here $Omega_{f}(H)$ is the subgroup of the $p$-group $H$ generated by all elements of order at most $p^{f}$. And the statement holds since, as noted above, if $G = langle x, y rangle$, then $G'$ is the normal closure of $langle [x, y] rangle$. If $[a, b]$ does not have order $p^{e}$, then $langle [x, y] rangle le Omega_{e-1}(G') < G'$, thus the normal closure of $langle [a, b] rangle$ is not $G'$, and so $a, b$ do not generate $G$.

Take $G=H_1oplus H_2$, where g.c.d.$(|H_1|,|H_2|)=1$, $phi_i: Fto H_i$.