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I was reading group theory from Topics in Algebra by I.N Herstein. I was studying about homomorphism. There was a theorem given which states that :

Let $\phi $ be a homomorphism of $G$ onto $\overline G$ with kernel $K$. Then $G/K\approx G$.

(Here $\approx$ stands for isomorphic, so $G/K\approx G$ means that $G/K$ is isomorphic to $G$ and the theorem is given as Theorem 2.7.1 in my version). A lemma was also introduced which stated that:

Suppose $G$ is a group, $N$ a normal subgroup of $G$; define the mapping $\phi$ from G to $G/N$ by $\phi (x) = Nx $ for all $x\in G$. Then $\phi$ is a homomorphism of $G$ onto $G / N$.

[It was given as Lemma 2.7.1 in my version]

However, after the (theorem stated above) there was a little detail added which goes like this:

Theorem 2. 7.1 is important, for it tells us precisely what groups can be expected to arise as homomorphic images of a given group. These must be expressible in the form $G/K$, where $K$ is normal in $G$. But, by Lemma 2. 7.1, for any normal subgroup $N$ of $G$, $G/N$ is a homomorphic image of $G$. Thus there is a one-to-one correspondence between homomorphic images of $G$ and normal subgroups of $G$.

However, I am not getting how are they concluding "Thus there is a one-to-one correspondence between homomorphic images of $G$ and normal subgroups of $G$." I mean I do get the fact that using the theorem I can check out all the groups which can be expected to appear as homomorphic images $G$ i.e those groups $\overline G$ will appear as a homomorphic image of $G$ which is isomorphic with $G/K$.

Now, since $\overline G$ is isomorphic with $G/K$ so, we can conclude that the group $G/K$ is somewhat equal to $\overline G$ as the difference lies only upon the labelling of the elements and knowing the group operation in $G/K$ we can perform an analogous group operation on $\overline G$, such that we also completely get to know $\overline G$. So, this is the reason why they say that the homomorphic images of $G$ say, $\overline G$ , can be expressed in the form $G/K$, where $K$ is a normal subgroup of $G$, since , $G/K\approx \overline G$ (or there exists an isomorphism from $G/K$ onto $\overline G$).

But I do not get how from Lemma 2.7.1 they are concluding, "Thus there is a one-to-one correspondence between homomorphic images of $G$ and normal subgroups of $G$."?

We know that $G/K$ is also a homomorphic image of $G$ and there may be other homomorphic images of $G$, as $\overline G$ depending upon if $G/K\approx \overline G$. For each normal subgroup $K$ of $G$ , we always get a homomorphic image $G/K$ of $G$. But now, there also may be several homomorphic images of $G$ different from the group $G/K$ depending upon the condition of the theorem above. So, for each normal subgroup $K$ we are getting several homomorphic images of $G$ (including $G/K$ and many others along with it $\overline G$ such that $G/K \approx\overline G$). So, how is there a one to one correspondance between $K$ and the homomorphic images of $G$ with respect to each such particular $K$? I am not quite getting it...

I actually perceive this differently, it goes like this :The reason for concluding is that each pair of isomorphic groups are somewhat equal as their difference only lies upon the labelling of the elements in the two groups considered. We can actually consider the two groups as two "sentences" written in two different languages and isomorphism as the dictionary and knowing the group operation in one group we can perform analogous operation in the other group as well and possibly conclude that the two sentences have the same meaning but they are just written in two different languages and the isomorphism and the group operation are actually our "keys" to decipher the sentence to the other language . So , if two groups let's say in the above case $G/K$ and $\overline G$ are isomorphic then $G/K$ and $G$ both are equal with symbols or labellings of each element changed in the case. Now, from the theorem since , the homomorphic image of $G$ is isomorphic with $G/K$ , that's how they are concluding it by considering that each of the isomorphic groups as one which is actually equal to $H/K$ for each $K$ and thus having a one-to-one correspondence with $K$. Can this be the reason?

But one more thing to note is that the homomorphic images of $G$ say, $\overline G$ has a one to one correspondance with $K$ if the homomorphism from $G$ to $\overline G$ is an onto mapping since, that is quoted in the theorem. If the homomorphism from $G$ to $\overline G$ is not onto , then the theorem fails to hold...so precisely , we can actually find the homomorphic images of $G$ say, $\overline G$ by this one to one correspondence for the onto homomorphisms of $G$ to $\overline G$ but we cant conclude the same i.e all homomorphisms of $G$ into $\overline G$ has a one to one correspondence with $K$ since all homomorphisms are not essentially onto. So , by this one to one correspondence we can find the onto homomorphisms of $G$ to $\overline G$ but not all homomorphisms. Is this valid?...

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1 Answer 1

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Here's a more explicit version of the theorem. Hopefully this helps clarify things.

Fix a group $G$.

First, let's define precisely how we'll talk about "homomorphic images of $G$".

The idea is look at pairs $(H, \ \varphi : G \to H)$, where $H$ is a group and $\varphi$ is a surjective group homomorphism $G \to H$ (this is the "witness" to the fact that $H$ is a homomorphic image of $G$). We say that $(H, \varphi)$ and $(H', \varphi')$ are isomorphic (as homomorphic images of $G$) exactly when

  1. There's an isomorphism $\alpha : H \cong H'$
  2. $\alpha \circ \varphi = \varphi'$ as maps $G \to H'$.

Notice that this notion of "isomorphism" reflects more than just the group structure of $H$! It also keeps track of the ~bonus structure~ that $H$ has -- namely, the structure of a surjection from $G$. As an aside, this kind of construction is quite common, and is a certain slice category under $G$ (some people would probably prefer I call it a coslice category, but that's beside the point). You'll almost certainly encounter this idea as you go further in math, and I encourage you to be on the lookout for it!

As a good exercise at this point, you should check that this notion of isomorphism really is an equivalence relation.


Now. Let $\mathcal{N}$ be the set of normal subgroups of $G$, and let $\mathcal{Q}$ be the set of homomorphic images of $G$ up to isomorphism (in the above sense).

There's a map $q : \mathcal{N} \to \mathcal{Q}$ which takes a normal subgroup $N$ and returns the equivalence class of $(G/N, \nu)$ where $\nu(g) = gN$ is the usual quotient map.

There's also a map $k : \mathcal{Q} \to \mathcal{N}$ which takes a homomorphic image $(H, \ \varphi : G \to H)$ and returns the normal subgroup $\text{Ker}(\varphi)$. (As another nice exercise, you should check this is well defined on equivalence classes!)

What do people mean when they say that "there is a one-to-one correspondence between normal subgroups of $G$ and homomorphic images of $G$"?

Theorem: $q$ and $k$ are mutually inverse, and thus provide a bijection between $\mathcal{N}$ and $\mathcal{Q}$.

As a last exercise, you should try to prove this yourself!


I hope this helps ^_^

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    $\begingroup$ Beat me to it, +1! $\endgroup$ Nov 24, 2022 at 5:35
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    $\begingroup$ One should also note that this precision is really needed. The imprecise version quoted in the question is wrong the way it's written: the quotients $G/N_1$ and $G/N_2$ may be isomorphic even though $N_1\neq N_2$, and then the one-to-one correspondence fails if we lump both quotients into the same equivalence class due to this isomorphism. Which is where @HallaSurvivor's notion of keeping track of the homomorphism is important, because then the quotients are no longer isomorphic. $\endgroup$ Nov 24, 2022 at 5:42
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    $\begingroup$ @HallaSurvivor Thank you! But , now, actually I perceive this differently, it goes like this :The reason for concluding is that each pair of isomorphic groups are somewhat equal as their differnce only lies upon the labelling of the elements in the two groups considered. We can actually consider the two groups as two "sentences" written in two different languages and isomorphism as the dictionary and knowing the group operation in one group we can perform analogous operation in the other group as well $\endgroup$
    – Arthur
    Nov 24, 2022 at 5:43
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    $\begingroup$ @Franklin -- exactly! If $\varphi : G \to \overline{G}$ is a surjective homomorphism, then there must exist a normal subgroup $K \vartriangleleft G$ so that $\overline{G} \cong G/K$. Indeed, we should take $K = \text{Ker}(\varphi)$. $\endgroup$ Nov 24, 2022 at 5:46
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    $\begingroup$ @Franklin -- Yup! This correspondence lets us totally understand the surjective homomorphisms out of $G$. This (by itself) doesn't tell us anything at all about non-surjective homomorphisms. But since every function is a surjection onto its image, if $\varphi : G \to H$ is any old homomorphism then we can write it as a composition $G \to \text{im}(\varphi) \to H$, where $G \to \text{im}(\varphi)$ is surjective, and $\text{im}(\varphi) \to H$ is injective. This lets us use this theorem about surjections to say something in more general settings $\endgroup$ Nov 24, 2022 at 6:06

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