Intuition with the Inverse Quotient Map

Question

Lemma

Suppose $$ J \subset \Gamma \subset G $$ where $J \triangleleft G$.

Then $\Gamma = \pi^{-1} (\Gamma /J)$. (This would not necessarily be true if $J \not \subset \Gamma$.)

So I understand that comment that if $J \not \subset \Gamma$, then since $ J \subset \pi^{-1} (\Gamma /J)$ this statement wouldn't be true.

However, I was struggling with the proof.

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Proof

It is trivial to show that $\Gamma \subset \pi^{-1} (\Gamma /J)$ since $\pi^{-1} (\Gamma /J) = \Gamma$.

Now we need to prove that $\Gamma \supset \pi^{-1} (\Gamma /J)$. We need to show $\; \pi(g) \in \Gamma /J \Rightarrow g \in \Gamma $.

Suppose $\pi(g) \in \Gamma /J$, then $\pi(g) = \gamma J $, where $\gamma J$ is some coset of $\Gamma/J$.

Then $$g \in \pi ^{-1}(\gamma J) \subset \gamma J \subset \Gamma J \subset \Gamma.$$

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I don't understand this step: $\pi ^{-1}( \gamma J) \subset \gamma J$.

Where does $\pi ^{-1} $ send cosets? I thought it would send the coset $\gamma J $ to $\gamma$.

Answer 1

Careful: $\pi^{-1}$ is not a function from $G/J$ to $G$; it cannot be, unless $\pi$ is a bijection.

Instead, $\pi^{-1}$ represents a function from the power set of $G/J$ to the power set of $G$: given a subset $A$ of $G/J$, $$\pi^{-1}(A) = \{g\in G\mid \pi(g)\in A\}.$$ By abuse of notation, if $A=\{\gamma J\}$ is a singleton, we write $\pi^{-1}(\gamma J)$ rather than $\pi^{-1}(\{\gamma J\})$.

That means that we cannot have $\gamma = \pi^{-1}(\gamma J)$. Certainly, $\gamma$ is in $\pi^{-1}(\gamma J)$, but unless $J$ is trivial, it will not be the only thing in $\pi^{-1}(\gamma J)$.

Now, for the proof that $\pi^{-1}(\Gamma/J)\subseteq \Gamma$, what I would do is the following: let $g\in\pi^{-1}(\Gamma/J)$. By definition, this means that $\pi(g)\in \Gamma/J$, so there exists $\gamma\in \Gamma$ such that $\pi(g) = \gamma J$. Thus, $gJ = \gamma J$, which means $\gamma^{-1}g\in J$. Since $J\subseteq \Gamma$ then $\gamma^{-1}g\in \Gamma$. Multiplying on the left by $\gamma$, we conclude that $g\in \Gamma$, as desired.

As to the argument you quote, it has two errors and elides a key step.

1. Error 1. It's incorrect to write "$\pi(g)\in\gamma J$". The correct statement is $\pi(g)=\gamma J$. Because $\pi(g)$ is the coset $gJ$, so it cannot be an element of the coset $\gamma J$, it is equal to the coset $\gamma J$.

2. Ommission of key step. The key step, that $\pi^{-1}(\gamma J)\subseteq \gamma J$ is just eliding the argument I gave above: suppose $g\in \pi^{-1}(\gamma J)$. Then $\pi(g) = \gamma J$, so $gJ=\gamma J$. This means that $g\in \gamma J$, since cosets are either identical or disjoint.

3. Error 2. Finally, note that "$\Gamma J \subseteq J$" is false unless $J=\Gamma$. Since $J\subseteq \Gamma$, you get $\Gamma J\subseteq \Gamma$ (in fact, equality), not $J$.