# Non-equivalence of groups and SESs of groups

There is a lemma in Lang's Algebra (2002, page 17): Let $H_1 \triangleleft G_1$, $H_2 \triangleleft G_2$, $f: G_1 \to G_2$, and $H_1 = f^{-1}(H_2)$. Then the homomorphism $$\hat f: G_1/H_1 \to G_2/H_2,\quad xH_1 \mapsto f(x)H_2$$ is well-defined.

I'm interested in the converse: if $\hat f$ is well-defined, do we necessarily have $H_1 = f^{-1}(H_2)$? I tried to prove it myself, but it seems that I need a hint.

My motivation is that if the converse holds, then the category with the objects of the form $\langle H, G\rangle$ such that $H \triangleleft G$ and the morphisms which are homomorphisms of the larger groups satisfying the condition from the lemma is equivalent to the category of short exact sequences of groups, and by extension it denotes the 'border' between the category of groups and category of short exact sequences of groups in the sense that it shows exactly why they are not equivalent.

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The answer is obviously no: let $f$ be the trivial homomorphism mapping everything to the identity element, and let $H_1$ be an arbitrary proper normal subgroup of $G_1$. Then $\hat{f}$ is still well-defined but $f^{-1}(H_2) = G_1 \ne H_1$. It is easily repaired, however: just require $\hat{f}$ to be injective. – Zhen Lin Jul 31 '11 at 8:33
Thanks, now I could easily prove the modified lemma, I'll gladly accept this as an answer ^_^ It's fascinating how Lang always barely mentions the most fundamental results in his book, teasing me :) – Alexei Averchenko Jul 31 '11 at 8:53
Glad I could help. (I am sometimes reluctant to post simple counterexamples as an answer out of fear of having misinterpreted the question. This has happened in the past.) – Zhen Lin Jul 31 '11 at 9:34

(Posting earlier comment as answer.)

The answer is obviously no: let $f : G_1 \to G_2$ be the trivial homomorphism mapping everything to the identity element, and let $H_1$ be an arbitrary proper normal subgroup of $G_1$. Then $\hat{f} : G_1 / H_1 \to G_2 / H_2$ is still well-defined but $f^{-1}[H_2] = G_1 \ne H_1$. Your claim is easily repaired, however: just require $\hat{f}$ to be injective.

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In order for the map to be well defined, a necessary condition is that $f(H_1)\subset H_2$, as this is what it would mean for the identity element in $G_1/H_1$ to be sent to the identity element in $G_2/H_2$. In fact, this condition is also sufficient: if $f(H_1)\subset f(H_2)$, then $f(gH_1)=f(g)f(H_1)\subset f(g)H_2$, so everything in the coset $[g]$ is sent to the coset $[f(g)]$, or more succinctly, $\widehat f([g])=[f(g)]$ is well defined.

If $H_1=f^{-1}(H_2)$, then since $\widehat f([g])=[e]\Leftrightarrow f(gH_1)=f(g)f(H_1)\subset f(g)H_2=H_2\Leftrightarrow f(g)\in H_2$ (note this holds without the condition, though the condition implies the containment is equality), we must have $f(g)\in H_2\Leftrightarrow g\in H_1$. Thus, $\widehat f$ is injective. If we run the same reasoning backwards, we see that $H_1=f^{-1}(H_2)$ if and only if $\widehat f$ is injective.

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Extending somewhat Aaron's answer:

Theorem. Let $G$ be a group, and let $N$ be a normal subgroup of $G$. If $f\colon G\to K$ is a homomorphism, then $f$ factors through the canonical projection $\pi\colon G\to G/N$ as $f = \widehat{f}\circ\pi$ (with $\widehat{f}(gN) = f(n)$ a homomorphism) if and only if $N\subseteq \mathrm{ker}f$. The map $\widehat{f}$ is one-to-one if and only if $N=\mathrm{ker}(f)$.

Proof. Suppose $f=\widehat{f}\circ\pi$ and $f$ is a homomorphism; then $N=\mathrm{ker}(\pi)\subseteq\mathrm{ker}(f)$ because the kernel of a composition of homomorphisms always contains the kernel of the first map applied. Conversely, suppose that $N\subseteq \mathrm{ker}(f)$. Then $\widehat{f}$ is well-defined: if $gN=g'N$, then $g^{-1}g'\in N\subseteq \mathrm{ker}(f)$, so $1=f(g^{-1}g')= f(g)^{-1}f(g')$, hence $f(g)=f(g')$, and $\widehat{f}$ is well-defined. It is also a group homomorphism, since $\widehat{f}(gg'N) = f(gg') = f(g)f(g') = \widehat{f}(gN)\widehat{f}(g'N)$. Finally, for every $g\in G$, $\widehat{f}\circ\pi(g) = \widehat{f}(gN) = f(gN)$, so $\widehat{f}\circ\pi = f$.

For the final clause: $\widehat{f}$ is one-to-one if and only if $\widehat{f}(gN) = 1$ holds if and only if $g\in N$; but $\widehat{f}(gN)=1$ if and only if $f(g)=1$ if and only if $g\in\mathrm{ker}(f)$, so $\widehat{f}$ is one-to-one if and only if $g\in N\Leftrightarrow g\in\mathrm{ker}(f)$, if and only if $N=\mathrm{ker}(f)$, as claimed. $\Box$

Your desired result now holds by letting $f = \pi_2\circ f_2$, where $\pi_2\colon G\to G/H_2$ is the canonical projection. The map factors through $G/H_1$ in the desired way if and only if $H_1\subseteq \mathrm{ker}(f)$, if and only if $H_1\subseteq \pi_2\circ f_2$, if and only if $H_1\subseteq f_2^{-1}(\mathrm{ker}(\pi_2))$, f and only if $H_1\subseteq f_2^{-1}(H_2)$, with equality if and only if the induced map is one-to-one.

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