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This is a question from a group theory exam. I have solved about 90% of the question. Actually, I have solved the whole question, but I want to solve it differently. There is no good reason to do it the the other way, except it feels like it should be easy and I can't stand being unable to solve it that way.

Let $G$ be a group and $H\preceq G$. Let $$f:G \rightarrow H$$ be a homomorphism such that the restriction $\left. f \right|_H$ is the identity $id_H$ on $H$.

For $a,\tilde{a} \in H$ and $b,\tilde{b} \in \text{ker}(f)$ show that:

a) If $H$ is normal then $ab=ba$.

Solution: If $H$ is normal then $$bab^{-1} \in H$$ for any $a \in H, b\in G$. It follows that $f(a) = a$ and $f(bab^{-1}) = bab^{-1}$. Because $f$ is a homomorphism and $f(b) = e_H$ we have that $$ bab^{-1} = f(bab^{-1}) = f(b) f(a) f(b^{-1}) = f(a) = a.$$ We conclude that $a = bab^{-1}$ as requested.

b) Explain why $$(a,b) \cdot (\tilde{a}, \tilde{b}) = (a\tilde{a}, b\tilde{b})$$ defines a group structure on $H \times \text{ker}(f)$.

Solution: This is just going trough the group axioms. I will spare you the details.

c) Let $\Psi: H \times \text{ker}(f) \rightarrow G$ be the map $$ \Psi(a,b) = ab.$$ Using part a), prove that $\Psi$ is an isomorphism.

Solution: It is not difficult to verify that $\Psi$ is a homomorphism. Furthermore, $\Psi$ is injective since if $\Psi(a,b) = \Psi(\tilde{a},\tilde{b})$ then $$ ab = \Psi(a,b) = \Psi(\tilde{a},\tilde{b}) = \tilde{a}\tilde{b}$$ so $$ ab = \tilde{a}\tilde{b}.$$ Applying $f$ we get $$ f(ab) = f(a) f(b) = a$$ and $$ f(\tilde{a}\tilde{b}) = \tilde{a}$$ So $a = \tilde{a}$. It follows automatically that $b = \tilde(b)$.

Problem: I can't find a proof that $\Psi$ is surjective. This is equivalent to proving that any $g\in G$ can be written $G = ab$ with $a \in H$ and $b \in \text{ker}(f)$. I can solve the question by constructing the inverse explicitly: $$ f^{-1}(g) = (f(g), f(g)^{-1}g).$$ But I don't want to. I want surjectivity :P.

edit: to be clear, I found the above inverse myself. I know it solves the question. But I want to show surjectivity without using the inverse. Actually, I'm just curious why every element $g \in G$ can be written $$ g = ab$$ with $a \in H$ and $b \in \text{ker}(f)$. This seems like a statement with content and the proof using the inverse does not help me understand it.

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  • $\begingroup$ Given a function $f: X\to Y$, constructing a function $u:Y\to X$ such that $f\circ u=id_Y$ is one way to prove surjectivity: every element $y\in Y$ is $f(x)$, with $x=u(y)$. $\endgroup$ – user228113 Apr 9 '16 at 10:49
  • $\begingroup$ @G.Sassatelli I understand, but I want to prove it directly. $\endgroup$ – Timon van der Berg Apr 9 '16 at 10:51
  • $\begingroup$ Perhaps you want a way to guess why the inverse is that one? $\endgroup$ – user228113 Apr 9 '16 at 10:52
  • $\begingroup$ Given $g\in G$, define $a=f\left(g\right)\in H$ and $b=f\left(g\right)^{-1}g$. It is easy to check that $b\in\ker f$, and $\Psi\left(a,b\right)=g$. $\endgroup$ – Guy Apr 9 '16 at 11:26
  • $\begingroup$ @guy Thanks, but the point of my question was: how do I show surjectivity without explicitly constructing the inverse? As you you can see, I had found the inverse myself. $\endgroup$ – Timon van der Berg Apr 9 '16 at 12:36
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For injectivity, since $\Psi$ is a homomorphism, you can just verify what it means that $\Psi(a,b)=1$. This means $ab=1$, so $a=b^{-1}\in\ker f$; therefore $a=f(a)=1$ and so also $b=a^{-1}=1$.

For surjectivity, consider $g\in G$. Then $a=f(g)\in H$, so $f(g)=a=f(a)$ and therefore $f(a^{-1}g)=1$. Hence, setting $b=a^{-1}g$, we have $$ g=aa^{-1}g=ab=\Psi(a,b) $$

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  • $\begingroup$ Very neat answer, but this does apparently not answer his question. This is exactly his method. $\endgroup$ – Jori Apr 10 '16 at 15:32
  • $\begingroup$ @Jori Where am I using $f^{-1}$? I think I'm exactly showing how the inverse can be built, rather than guessed at. $\endgroup$ – egreg Apr 10 '16 at 15:40
  • $\begingroup$ Fair enough. The question is rather vague anyhow. In general I think it is hard to say whether two proofs are similar or not. $\endgroup$ – Jori Apr 10 '16 at 15:54
  • $\begingroup$ @Jori And there's not much more to do: the information is that $f(g)\in H$, so $f(g)=f(f(g))$ and therefore $f(g)^{-1}g\in\ker f$. After all, the inverse map is unique! ;-) $\endgroup$ – egreg Apr 10 '16 at 16:09
  • $\begingroup$ Sure, +1. My method is slightly different, but it boils down to the same things of course. For the folks (like me) that did not realize how easy it was to construct a specific (simple) inverse function, haha. $\endgroup$ – Jori Apr 10 '16 at 16:15
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Maybe you are looking for something in which the properties of $f, G, H$ are more specifically used and the proof follows more naturally (but your approach is certainly valid, did you find it by trial and error?):

We must show that for any $g \in G$ we have for some $a \in H$ and $b \in \ker(f)$ such that $g = ab$. Because $\ker(f)$ is a subgroup of $G$ we know that we can write $g = \hat{g}b'$ for some $\hat{g} \in G$ and $b' \in \ker(f)$. Also since $H$ is normal in $G$ we can rewrite $a$ to $\hat{g}a'\hat{g}^{-1}$ for some $a' \in H$. Therefore we get that we must show that $\hat{g}a'\hat{g}^{-1}b = \hat{g}b' \Leftrightarrow a'\hat{g}^{-1}b = b'$. Remember that we can choose $a, b$ freely and therefore because of bijection also $a'$. If we could show now that $a'\hat{g}^{-1} \in \ker(f)$ for some $a'$ we would be done (again because of the bijection we create by multiplying freely with $b$ and remembering that $\ker(f)$ is a subgroup). This last statement is obvious because $f$ maps $G$ into $H$, and we are done

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  • $\begingroup$ This is what I was looking for, but I accepted egrerg's answer since it is slightly more elegant (though he does not emphasize the part I am interested in, it is there). Now that I am typing this, I think my own proof contains the same information. I just did not realize that it did. To answer your question: I'm working on that problem as well. Let me know if you find anything. $\endgroup$ – Timon van der Berg Apr 10 '16 at 17:09

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