How to prove that a set is universe of a subgroup?

Question

I'm trying to solve this proposition: Let $ \langle A, \cdot, ^{-1}, 1\rangle $ and $ \langle B, \cdot, ^{-1}, 1\rangle $ be groups and let $ \alpha \colon A \to B $ be a homomorphism. Then the set $ N \mathrel{\mathop:}= \{ a \in A : \alpha(a) = 1 \} $ is the universe of a subgroup of A.

I've trying to let a $ X \subseteq A $ such that $ X $ is a subgroup of A. After that, I get an element of $ X $, say $ a \in X $ and trying to get $ \alpha(a) = 1 $ through $ a \in A $ and $ 1 \in A, B, X $. But no idea how to continue the demonstration

Proposition image

Answer 1

There is no need to define a set $X$. Instead, what you need to show is that:

1. The restriction of $\cdot$ to $N\times N$ is a binary operation on $N$;
2. The restriction of ${}^{-1}$ to $N$ is a unary operation on $N$;
3. The co-restriction of $1$ to $N$ is still well-defined (that is, the image lies in $N$;
4. The group identities are satisfied when all operations are restricted as above.

(4 may be moot, or trivial). Once you do that, you will have that $\langle N, \cdot|_{N\times N},{}^{-1}|_N,1\rangle$ is a group (hence $N$ is the universe/underlying set of a group), and since $N\subseteq A$ by construction, it is in fact a subgroup of $\langle A,\cdot,{}^{-1},1\rangle$.