TITLE: If two finite groups satisfy the same first-order sentences, are they isomorphic?
QUESTION [5 upvotes]: My question is the title. I would be glad if someone could supply a proof if true, or a counterexample if false.

REPLY [3 votes]: Let us spell out an example of Andre's answer (for concreteness).
Suppose a group $G$ satisfies the same first-order sentences as $C_2$, the cyclic group of order $2$.

Then $G$ satisfies the following sentence, asserting that there are at most $2$ elements in the group. $$\forall xyz(x=y \vee y=z \vee x=z)$$
And $G$ satisfies the following sentence, which encodes the diagram of $C_2$ (also known as its Cayley table) as a first-order formula.

$$\exists xy(x \neq y \wedge xx=x \wedge xy=y \wedge yx=y \wedge yy=x)$$
But this means that $G$ is isomorphic to $C_2$.