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Clearly the kernel of a group homomorphism is normal (proof), but I often hear my professor mention that any normal subgroup is the kernel of some homomorphism.

This feels correct but isn't entirely obvious to me.


One thought I had is that for any normal subgroup $N$ of $G$, we could define the quotient homomorphism $\pi:G\to G/N$ since $G/N$ is a group.

I was imagining that we could consider $\pi^{-1}:G/N\to G$, whose kernel would then be $N$. However, $\pi^{-1}$ doesn't exist since $\pi$ is not a bijection in general.


So my question is this: is there an obvious way to define a homomorphism whose kernel is an arbitrary normal subgroup of $G$? Or does it depend on the particular group whether you can define such a homomorphism?

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    $\begingroup$ No, you more or less have it, the kernel of $\pi\colon G\to G/N$ is $N$. No need to worry about $\pi^{-1}$, or if it exists. $\endgroup$
    – Ben West
    Jul 18, 2015 at 23:17

3 Answers 3

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You only have to consider that $\pi:G\to G/N$, defined by $\pi(x)=xN$, is a homomorphism and the $\ker \pi$ is precisely $N$.

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    $\begingroup$ overcomplicating as usual. thanks. $\endgroup$
    – pancini
    Jul 18, 2015 at 23:18
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    $\begingroup$ that's why we are meant to be here $\endgroup$
    – janmarqz
    Jul 18, 2015 at 23:19
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    $\begingroup$ What does it mean exactly to have $\pi(x)=xN$? Shouldn't $\pi(x)$ yield a group element of $G/N$? If it should, why does $x$ act on the whole subgroup $N$, and not just a specific element of $N$? $\endgroup$
    – zabop
    Dec 10, 2022 at 23:20
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    $\begingroup$ @zabop, $xN$ is the (left) coset of $N$ of representative $x\in G$, so it is precisely an element of the quotient group $G/N$. $\endgroup$
    – Kan't
    Dec 11, 2022 at 7:09
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    $\begingroup$ Oh, the elements of the quotient group are cosets of the original group; nice, I missed this. Thanks. $\endgroup$
    – zabop
    Dec 11, 2022 at 9:44
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Anyway, if the kernel is non-trivial, the homomorphism is not injective!

You're confusing the inverse of an isomorphism, and the inverse image of a subset by a map. So, yes, if $N$ is a normal subgroup, it is the kernel of the canonical homomorphism : \begin{align*}\pi\colon G&\longrightarrow G/N,\\ g&\longmapsto gN. \end{align*} Indeed, $\pi^{-1}(\bar 1)=\pi^{-1}(N)=N$.

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Your intuition is correct, and unfortunately, yes, $\pi^{-1}$ isn't a function. Could we somehow "make" it one?

Well, one way this is often done with "ordinary" functions, is to treat $f^{-1}$, for a given function $f: A \to B$, as not something like this:

$f^{-1}: B \to A$

but instead considering $f^{-1}$ as a function between power sets:

$f^{-1}: \mathcal{P}(B) \to \mathcal{P}(A)$

where for a subset $Y \subseteq B$, we define $f^{-1}(Y) = X = \{x \in A: f(x) \in Y\}$.

Applying this definition to $\pi^{-1}$, we get:

$\pi^{-1}(e_{G/N}) = \pi^{-1}(N) = \{g \in G: \pi(g) \in N\}$

$= \{g \in G: gN = N\} = \{g \in G: ge^{-1} \in N\} = \{g \in G: g \in N\} = N$

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  • $\begingroup$ The definition for a different object deserves a different symbol: let me propose $f^{\leftarrow}(Y)$ for your $f^{-1}(Y)$. $\endgroup$
    – Kan't
    Dec 11, 2022 at 8:23

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