So, for each of the following groups write a formula in the language of group theory, which holds in given group, but doesn't hold in others two.
$(i)$ The integers with addition \ I think it's $\{\Bbb{Z},+,-,0\}$
$(ii)$ Positive reals with multiplication \ I think it's $\{\Bbb{R}_+,*,\,^{-1},1\}$
$(iii)$ Permutations of $\{a,b,c\}$ with composition \ I don't know exatly what it is.
So, I found such formula for $(ii)$, it's:
$\forall x \exists y(x=y*y)$;
This formula holds for $(ii)$ but doesn't hold for $(i)$, 'cause $\forall x\exists y(x=y*y)$ to $(i)$ is $\forall x\exists y(x=y+y)$, doesn't hold for $x=1$.
But I've stuck on others groups and I cannot fid anything right for them. Notice that we don't have '$>$' and '$<$' for the first two, and I'm not sure about last one.
If you know anything about group theory, $(iii)$ is just $S_3$ and has exactly $6$ elements (which you can write in the language of groups since you have equality). Clearly, "Having exactly 6 elements" is true in $(iii)$ and false in $(i)$ and $(ii)$. We call this sentence $\varphi$.
As you noted, $\mathbb{R}_+ \models \forall x \exists y (x = y*y)$. This is false in $(i)$ and it is also false in $(iii)$. Notice that a transposition, i.e. $(ab)$, cannot be written as the composition of two permutations. We call this sentence $\psi$.
Finally, as Derek Holt suggested, we are already done. Notice that $\neg \varphi \wedge \neg \psi$ is a sentence in FOL which does not hold in $(ii)$ (since $\psi$ holds in $(ii)$) and does not hold in $(iii)$ (since $\varphi$ holds in $(iii)$), but does hold in $(i)$ because $\neg \psi$ and $\neg \varphi$ hold in $(i)$.