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

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

• You don't have to "let $X$ be" anything. You want to show that $N$, which you have defined, is a subgroup. To start, can you show that if two things are in $N$ then so is their product? – Ethan Bolker Jun 4 at 21:58
• The universe of a subgroup? What is that thing? Everything you wrote is extremely unclear. Do you want to simply show that $N$ (a.k.a. the kernel of $\alpha$) is a subgroup of $A$? – freakish Jun 4 at 22:29
• The proposition says "[...] N is the universe of a subgroup of A", isn't clear for myself too, I'm not allow to edit and put a picture of the proposition, but I'm hosting a image of that and put on the text – itepifanio Jun 4 at 22:38
• Here's my best guess regarding the usage of terminology universe, based on the notation "$<A, \cdot, ^{-1}, 1\ >$" of the exercise. That notation hints to me that in the context of this exercise a group is required to be expressed very formally as a 4-tuple. Even a subgroup is required to be a 4-tuple. The 4th entry of that 4-tuple is the identity, the third element is the inversion operator, the second element is the group operation, and the first element is the "universe" of the group; the terminology that I've heard more commonly would be the "underlying set" of the group. – Lee Mosher Jun 4 at 22:46

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 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 $$\langle A,\cdot,{}^{-1},1\rangle$$.