4
$\begingroup$

In texts which present the third isomorphism theorem: $$(G/N)/(H/N) \cong G/H$$

the relationship between the entities is often seen presented in the form:

Let $H$ and $N$ be normal subgroups of a group $G$, and $N \subseteq H$. Then $H/N$ is a normal subgroup of $G/N$ and there is an isomorphism $(G/N)/(H/N) \cong G/H$.

(EDIT: This is how it is presented in Allan Clark's Elements of Abstract Algebra (1971), article $68$, for example.)

I have been in discussion with someone who insists that it is mandatory to state that $N$ is a normal subgroup of $H$, or the definition is invalid, because if $N$ is not a normal subgroup of $H$ then $H/N$ is undefined.

My own personal preference is that the statement of the theorem is perfectly good as it stands, because if $N$ and $H$ are both normal in $G$ and $N \subseteq H$, then it logically follows that $N$ is a normal subgroup of $H$ and it is therefore unnecessary and inelegant to specifically state this in the exposition.

What one would of course do is to derive the property during the course of the proof, or reference it as a pre-proven lemma.

Indeed, we have, by definition of normal subgroup: $$\forall g \in G: g H = H g$$

$$\forall g \in G: g N = N g$$

and so:

$$\forall g \in H: g N = N g$$

But, it is argued, the above is non-trivial and far from obvious, and therefore it is important in the exposition to make that explicit statement: "Let $N$ be a normal subgroup of $H$."

To my mind it is better to merely state subsetness, and to prove normal-subgroupness during the course of the proof itself. Otherwise it may leave the reader open to ask the question: "But what happens if $N$ is a subset of $H$ but not necessarily a normal subgroup of $H$? What happens then?" without taking that mental step of realisation that you can't have $N$ being a subset of $H$ under the above circumstances without $N$ being a normal subgroup of $H$.

In other words, I believe it important to establish that (on the surface) "more general" statement of the problem, and it is unhelpful to explicitly include one of the facts that is established during the course of the proof.

What is the general consensus of opinion here?

$\endgroup$
2
  • 2
    $\begingroup$ I think that it would be best to prove this fact about the normality of $N$ in $H$ earlier. (It is a natural thing to prove or have as an exercise.) Then put in a line immediately before or after the theorem "Recall that..." to remind the reader that they don't have to worry about $H/N$ being merely a coset space. $\endgroup$
    – Adam
    Jul 19, 2022 at 12:34
  • 3
    $\begingroup$ So your friend definitely insists on treating it as another (redundant) condition? Because it's pretty easy to just state that $N$ is a normal subgroup of $H$ while making it clear that it's a consequence of the previous conditions. Perhaps there was a miscommunication somewhere? $\endgroup$
    – Thierry
    Jul 19, 2022 at 17:05

4 Answers 4

7
$\begingroup$

I agree that $\text“\color{red}{N \text{ is a normal subgroup of a group }H}\text”$ is a redundant/bogus condition of the given theorem.

  1. To specify what is rightly a $\color{red}{\textrm{lemma}}$ of the third isomorphism theorem as one of its conditions instead would, at the logical level—unnecessarily—weaken the theorem, since $$∀g{,}h{,}n\Big(Aghn→Bghn\Big) \quad\implies\quad ∀g{,}h{,}n\Big((Aghn∧\color{red}{Lhn})→Bghn\Big)$$ while $$∀g{,}h{,}n\Big((Aghn∧\color{red}{Lhn})→Bghn\Big) \quad\kern.6em\not\kern-.6em\implies\quad ∀g{,}h{,}n\Big(Aghn→Bghn\Big).$$

  2. Echoing your sentiment: thoughtful readers whose legitimate proof derives that lemma instead of invoking it by assumption, will be needlessly wondering whether their work and understanding is indeed rigorous and valid, and will be futilely trying to figure out where that condition is required.

If one wishes to highlight that lemma within the statement of the third isomorphism theorem, then, while it's not illogical to specify it as a condition, it is more honest, more helpful, and ultimately cleaner to specify it as a consequence instead. Some suggestions:

  • Let $H$ and $N$ be normal subgroups of a group $G,$ and $N \subseteq H,$ so that $N$ is a normal subgroup of a group $H.$

    Then $H/N$ is a normal subgroup of $G/N,$ and there is an isomorphism $(G/N)/(H/N) \cong G/H.$

  • Let $H$ and $N$ be normal subgroups of a group $G,$ and $N \subseteq H;$ so, $N$ is a normal subgroup of a group $H.$

    Then $H/N$ is a normal subgroup of $G/N,$ and there is an isomorphism $(G/N)/(H/N) \cong G/H.$

  • Let $H$ and $N$ be normal subgroups of a group $G,$ and $N \subseteq H.$ (So, $N$ is a normal subgroup of a group $H.)$

    Then $H/N$ is a normal subgroup of $G/N,$ and there is an isomorphism $(G/N)/(H/N) \cong G/H.$

  • Let $H$ and $N$ be normal subgroups of a group $G,$ and $N \subseteq H.$

    Then $N$ is a normal subgroup of a group $H,$ $H/N$ is a normal subgroup of $G/N,$ and there is an isomorphism $(G/N)/(H/N) \cong G/H.$

$\endgroup$
1
  • 5
    $\begingroup$ The last of those options works best for me. Thanks. $\endgroup$ Jul 19, 2022 at 16:08
4
$\begingroup$

I would definitely want to state that $N$ is a normal subgroup of $H$ before saying anything that treats $H/N$ as a well-defined group. That statement of normality could reasonably be part of the conclusion (not the hypothesis) of the theorem, or it could be presented earlier in the text (perhaps as an exercise, since it's pretty trivial).

I would also explicitly state, perhaps as part of the theorem or perhaps as an addendum after the proof, the explicit form of the isomorphism. The explicit isomorphism, not just the existence of some isomorphism, is important.

$\endgroup$
3
$\begingroup$

I also agree with ryang that stating the normal subgroup status of $N$ in $H$ as a condition would be weakening the theorem. However, explicating and demonstrating that fact somewhere in the vicinity of the proof seems like good form.

Here's the theorem in Hungerford's Abstract Algebra: An Introduction. Note that, like the OP's referenced text, it doesn't include the normal-subgroup status of $N$ in $K$ explicitly in this statement:

THEOREM 7.30 (THIRD ISOMORPHISM THEOREM FOR GROUPS) Let $K$ and $N$ be normal subgroups of a group $G$ with $N \subseteq K \subseteq G$. Then $K/N$ is a normal subgroup of $G/N$, and the quotient group $(G/N)/(K/N)$ is isomorphic to $G/K$.

However, the fact is given in prelude text just before the theorem:

Let $N$ be normal subgroup of a group $G$... Let $K$ be a subgroup of $G$ that contains $N$. Then $N$ is certainly a subgroup of $K$. Since $Na = aN$ for every $a \in G$, then in particular $Na = aN$ for every $a \in K$. Hence $N$ is a normal subgroup of $K$, so that $K/N$ is a group by Theorem 7.23.

This seems good to me.

$\endgroup$
3
$\begingroup$

When using expressions whose meaning depends on properties, you should proclaim those properties before using the expression. Here something of the form $G/H$ is generally a set of cosets, but if $H$ is a normal subgroup of $G$ then one can (and usually does) use the same expression to denote a group. Since one can only form the quotient by a (normal) subgroup (at least if the result is to be considered as a group), it is problematic to interpret $(G/N)/(H/N)$ unless it it known that $H/N$ denotes a group (and indeed a normal subgroup of $G/N$). I think the following formulation should satisfy even the pedantically precise:

Let $H$ and $N$ be normal subgroups of a group $G$, and $N \subseteq H$. Then $N$ is a normal subgroup of $H$ so that $H/N$ is a group, and this group is (naturally identified with) a normal subgroup of the quotient group $G/N$, and the rule $gH\mapsto (gN)(H/N)$ for $g\in G$ determines a well defined map $G/H\to (G/N)/(H/N)$ that is an isomorphism of groups.

I've added the (naturally identified with) because from some points of view a quotient group constructs a group in a fundamentally different way than a subgroup (the former comes with a canonical projection morphism, the latter with an embedding morphism, for instance), though in this case the set-theoretic nuts and bolts that make up a group proper (underlying set, identity and inverse maps) are exactly the same for the two groups in question.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.