# Is there a nice way to use the group isomorphism theorems in this proof?

The following theorem is an exercise in the section of a textbook dealing with the group isomorphism theorems. However, I did not use the isomorphism theorems to prove this theorem, so I wonder if there might be a cleaner way using them?

Theorem: Let $$(A, \cdot)$$, $$(B, \ast)$$ be groups, let $$f: A \mapsto B$$ be a homomorphism, and let $$S^\ast \leqslant B$$. Then: $$\textrm{Ker}(f) \leqslant f^{-1}(S^\ast) = \{x \in A \mid f(x) \in S^\ast\} \leqslant A$$

Proof. Let $$x, y \in f^{-1}(S^\ast)$$. Then, we know that there exists elements $$f(x), f(y) \in S^\ast$$. Since $$S^\ast \leqslant B$$, $$f(x) \ast f(y) \in S^\ast$$. Since $$f$$ is a homomorphism, $$f(x) \ast f(y) = f(x \cdot y) \in S^\ast$$. This means that $$x \cdot y \in f^{-1}(S^\ast)$$, so $$f^{-1}(S^\ast)$$ is closed under the group operation.

Since $$S^\ast$$ is a subgroup, we know that if $$f(x) \in S^\ast$$ (where $$x \in f^{-1}(S^\ast)$$), then there exists an element $$f(y) \in S^\ast$$ (where $$y \in f^{-1}(S^\ast)$$), such that $$f(x) \ast f(y) = \textrm{id}_{B}$$. Since $$f$$ is a homomorphism, it must be that $$y = x^{-1}$$. Thus, $$x^{-1} \in f^{-1}(S^\ast)$$.

Thus, we have shown that $$f^{-1}(S^\ast)$$, and every element in it has its inverse within the set too, so $$f^{-1}(S^\ast)$$ is a subgroup of $$A$$. It remains to show that $$\textrm{Ker}(f) \subset f^{-1}(S^\ast)$$. Since $$S^\ast$$ is a group, $$\textrm{id}_{B} \in S^\ast$$. Therefore, $$\textrm{Ker}(f) = f^{-1}(\textrm{id}_{B}) \subset f^{-1}(S^\ast)$$.

In the second part, you say that since $$f(x) \in S^*$$ then there must be $$f(y) \in S^*$$ such that $$f(x)*f(y)=1_B$$. Actually I would only deduce that there is $$v \in S^*$$ such that $$f(x)*v=1_B$$. And maybe I can guess your doubts about surjectivity came from here.
A way to make it more understandable (at least to me!) is the following. Let $$x \in f^{-1}(S^*)$$, so that as you said $$f(x) \in S^*$$. Take $$y=x^{-1}$$. You have to prove that $$y \in f^{-1}(S^*)$$, i.e. $$f(y) \in S^*$$. But $$f(x)*f(y)=f(x \cdot y)=f(1_A)=1_B$$ so that $$f(y)$$ is the inverse of $$f(x)$$ in $$B$$. Since $$S^*$$ is a subgroup, it is close for taking inverses and we are done.