Model Theory: Group Substructures

Question

Maria Manzano defines a group in two different ways:

1. $\langle A, e, + \rangle$ in which $e$ is the neutral element, $+$ is associative, and each element has an inverse.
2. $\langle A, e, \space^{-1}, + \rangle$ in which $e$ is the neutral element, $+$ is associative, and for each $x\in A$, $x+x^{-1}=e$.

The question is this: Is any substructure of a group, a group? I think if you consider the first definition, the answer could be no, but if you consider the second it would be yes, but the author says the second definition is an alternative to the first. Am I wrong?

Answer 1

You are right : in the first definition, you may have substructures that are not themselves groups. An example is $(\mathbb N,+,0)$ sitting in $(\mathbb Z,+,0)$ : it is clearly a substructure but not a group.

The two definitions are equivalent in a looser sense : the structures are the same, but the notion of substructure will not be the same.

I don't know which book you're reading from, but if it's about group theory, it won't matter because people will always say "subgroup" rather than substructure. The difference comes into play if you're reading something about model theory or universal algebra.

Answer 2

Given a $\langle A,e,+\rangle$ as described, the inverse that is guaranteed to exist is necessarily unique. That is, If $x+y=e$ and also $x+z=e$, then $$\tag1y=y+e=y+(x+z)=(y+x)+z=e+z=z.$$ Hence we can defin a map ${}^{-1}\colon A\to A$ that sends each $x\in A$ to its uniquely determined inverse. This allows us to extend $\langle A,e,+\rangle$ in a unique way to $\langle A,e,{}^{-1},+\rangle$, i.e., the second kind of structure turns out to be equivalent to the first kind.

If we could not use $(1)$, your doubts would be valid.