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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

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    $\begingroup$ 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? $\endgroup$ – Ethan Bolker Jun 4 at 21:58
  • $\begingroup$ 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 $A$? $\endgroup$ – freakish Jun 4 at 22:29
  • $\begingroup$ 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 $\endgroup$ – itepifanio Jun 4 at 22:38
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    $\begingroup$ 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. $\endgroup$ – Lee Mosher Jun 4 at 22:46
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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 $\langle A,\cdot,{}^{-1},1\rangle$.

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