2
$\begingroup$

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, $\textrm{Ker}(f) = f^{-1}(\textrm{id}_{B}) \subset f^{-1}(S^\ast)$.

$\endgroup$
1
$\begingroup$

I don't believe the isomorphism theorems are necessary in order to prove the theorem. Your proof is the one I would use. Just because the exercise happens to appear in a section dealing with isomorphism theorems does not mean that isomorphism theorems are necessary for the proof. Now, if the exercise had specifically mentioned to use the isomorphism theorems in the proof, that would be different, but again I don't see how the isomorphism theorems would be helpful here.

$\endgroup$
1
$\begingroup$

I think there is something which is not totally clear in this proof. As for the first part, everything is ok.

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.

Sorry if I wrote so much, I tried to be as clear as I could :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.