Understanding quotient groups and normal subgroups, intuitively

Jan 2021

One of the defining characteristics of math is breaking down complex objects into simpler pieces that are easier to study. For instance, we prime-factorize integers and decompose matrices, and in the process gain a better understanding of the very nature of our original objects. In group theory where the objects of study are groups, this question becomes: how do we break down complicated groups into more wieldy ones?


After learning about Lagrange’s theorem and how subgroups partition groups into the original subgroup and its cosets, a natural idea might be to leverage this to break down groups. For example, take the subgroup H={0,2}H = \{0, 2\} of Z4\mathbb Z_4 and its coset 1+H={1,3}1 + H = \{1, 3\}. By reorganizing the Cayley tables, we might end up with something like this:

Z4 and its cosets

If you squint, you might see something that resembles the Cayley table of Z2\mathbb Z_2, where instead of the integers 00 and 11 we have the cosets 0+H=H0 + H = H and 1+H1 + H. And just based off the clumped Cayley table, there’s a sensible operation * to “combine” these cosets to give another coset:

(a+H)(b+H)={x+y:xa+H,yb+H}=(a+b)+H. \begin{aligned} (a+H) * (b+H) &= \{x + y : x \in a+H, y \in b+H\} \\ &= (a+b) + H. \end{aligned}

=(a+H)(b+H)={x+y:xa+H,yb+H}=(a+b)+H. \begin{aligned} &\phantom{=} (a+H) * (b+H) \\ &= \{x + y : x \in a+H, y \in b+H\} \\ &= (a+b) + H. \end{aligned}

For instance, H(1+H)={0+1,0+3,2+1,2+3}={1,3}=1+HH * (1+H) = \{0 + 1, 0 + 3, 2 + 1, 2 + 3\} = \{1, 3\} = 1+H. And it even seems to work when we choose a different coset representative 2+H=H2 + H = H, since (2+H)(1+H)=1+H(2+H) * (1+H) = 1+H as well. It’s a great exercise to verify this works for any combination of HH and 1+H1+H, and with different coset representatives. So now we’ve got a set of two elements, {H,1+H}\{H, 1+H\} that has a closed operation. It looks like a group and smells like one1, one smaller than the group Z4\mathbb Z_4 of four elements we started with!

Daringly, you might try this process on a nonabelian group, with the subgroup H={(),(123),(132)}H = \{(), (123), (132)\} of S3S_3 and its coset (12)H={(12),(23),(13)}(12)H = \{(12), (23), (13)\}. Again, something resembling Z2\mathbb Z_2 seems to appear:

S3 with cosets of A3

Similarly, we can combine these cosets with * to produce another coset2:

aHbH={xy:xaH,ybH}=abH. \begin{aligned} aH * bH &= \{xy : x \in aH, y \in bH\} \\ &= abH. \end{aligned}

=aHbH={xy:xaH,ybH}=abH. \begin{aligned} &\phantom{=} aH * bH \\ &= \{xy : x \in aH, y \in bH\} \\ &= abH. \end{aligned}

So again the set {H,(12)H}\{H, (12)H\} is equipped with a closed operation *, making it a very convincing group3 and three times smaller than our original S3S_3! You might start to wonder if this process works for all groups. That’s until you try a second nonabelian example for good measure, with H={(),(12)}H = \{(), (12)\} of S3S_3 and its cosets (123)H={(123),(13)}(123)H = \{(123), (13)\} and (132)H={(132),(23)}(132)H = \{(132), (23)\}:

S3 without clumping

This time, the cosets didn’t form tidy clumps! In particular, if we take HH and try to combine it with (123)H(123)H, we find that

H(123)H={xy:xH,y(123)H}={(123),(13),(23),(132)}. \begin{aligned} H * (123)H &= \{xy : x \in H, y \in (123)H\} \\ &= \{(123), (13), (23), (132)\}. \end{aligned}

=H(123)H={xy:xH,y(123)H}={(123),(13),(23),(132)}. \begin{aligned} &\phantom{=} H * (123)H \\ &= \{xy : x \in H, y \in (123)H\} \\ &= \{(123), (13), (23), (132)\}. \end{aligned}

But {(123),(13),(23),(132)}\{(123), (13), (23), (132)\} is not a coset itself, much less equal to (123)H(123)H! Sadly, with this particular subgroup H={(),(12)}H = \{(), (12)\}, the general rule aHbH=abHaH * bH = abH breaks. The next question then becomes: what made the first two subgroups special?

Building Intuition

Before we answer that question, I want to present some intuition that’ll guide us in arriving at the answer–an answer that always felt a little arbitrary to me otherwise.

First of all, what does it mean for two elements to be in the same coset? Well, elements x,yx, y land in the same coset yHyH exactly when x=yhfor some hH.x = yh \quad\quad\text{for some } h \in H. On an intuitive level, you might interpret this as xx is “yy-like”. In our original group, we would have differentiated between xx and yy. But once placed into the same coset, we agree that xx is “just enough like” yy such that they end up in the same coset. And that loss of differentiation is a tradeoff: we lose some detail of our original group but end with a smaller one–like lowering the resolution on an image.

To solidify this intuition, let’s define “yy-like.”

DEFINITION. An element xx is said to be “yy-like” if and only if x=yhx = yh for some hHh \in H.

Now let’s return to aHbH=abHaH * bH = abH. This was our previous requirement to create a smaller group from the cosets, and with this new definition, it should feel natural. We’d expect an element of aHbHaH * bH that looks like an aa-like element multiplied by a bb-like element to produce an abab-like one in abHabH. Conversely, we’d expect an abab-like element in abHabH to be the product of an aa-like and bb-like element in aHbHaH * bH. Overall, we’d expect aHbH=abHaH * bH = abH.

We saw that not all groups satisfy this requirement and so don’t create a group from the cosets. But this requirement is a little hard to work with since it requires us to check if all combinations of the cosets, and when they have different coset representatives. Instead, it’d be great if there was some logically simpler condition that guaranteed aHbH=abHaH * bH = abH for free.

Let’s see where our intuition can take us. We know that when we left-multiply some element gg by any hHh \in H, we end up with an element that is gg-like. You might notice this sounds an awful lot like how the identity works: left-multiplying an element gg by the identity gives you back gg. But in hh’s case, it doesn’t give you back exactly gg, but something gg-like. So elements hHh \in H are identity-like!

Of course, the usual identity is double-sided. If elements of HH are truly identity-like, we might expect them to fulfill this duty faithfully. So we’d like that if we right-multiply gg by some hHh \in H, that produces something gg-like as well. In other words, if we left-multiply gg by every hHh \in H to get all gg-like elements, they better be the same elements we get from right-multiplying gg by every hHh \in H. In the language of cosets, this condition becomes gH=Hgfor all gG.gH = Hg \quad\text{for all } g \in G. At a glance, this condition4 seems a lot simpler to work with since we don’t have to check combinations of different cosets. And as a it turns out, the coset operation aHbH=abHaH * bH = abH is well-defined if and only if this condition gH=HggH = Hg for all gGg \in G holds5.

Because of their importance, we call subgroups that satisfy this condition normal subgroups in the sense that they are the only subgroups that allow us to form a group out of the cosets. And we call this group of the cosets the quotient group G/H={gH:gG}G/H = \{gH : g \in G\} (read GG mod HH).

DEFINITION. A subgroup HH is normal if and only if gH=HggH = Hg for all gGg \in G.

DEFINITION. A quotient group is the set of cosets G/H={gH:gG}G/H = \{gH : g \in G\} with the operation * defined by aHbH=abHaH * bH = abH for all a,bGa, b \in G, where HH acts as the identity element and each coset has an inverse, namely g1Hg^{-1}H.

When I first learned about normal subgroups, they were special subgroups that satisfied gHg1=HgHg^{-1} = H. While a logically equivalent statement (and easier to work with when it comes to showing a specific subgroup is normal), it just felt like some random symbolic coincidence. But by observing that HH ought to act like the identity, the equivalent condition gH=HggH = Hg became more natural. Further, this intuition of HH being identity-like extends well to an overarching understanding of quotient groups. You’re breaking down a group by modding/quotienting out the information that you consider to be “just enough like” the identity–i.e. information and detail you don’t care about!

In the first example above, we quotiented Z4\mathbb Z_4 by H={0,2}H = \{0, 2\}. We considered 22 to be “just enough like” 00 because what differentiates 00 and 22 in Z4\mathbb Z_4 is irrelevant to us in Z/{0,2}\mathbb Z/\{0, 2\}. Both 00 and 22 are even, so all we care about is parity, which is also why the quotient group resembles Z2\mathbb Z_2. In the second example, we didn’t care about the specific permutations, but again just their parity by considering even-transposition permutations (123)(123) and (132)(132) “just enough like” ()(). As a result, the quotient group resembles Z2\mathbb Z_2 with cosets A3A_3 and (12)(12)A3A_3.

The First Isomorphism Theorem

Equipped with these ideas, we can now tackle an important theorem that deals with quotient groups, called the First Isomorphism Theorem. But first, there’s a bit of terminology to introduce. Our process of clumping elements into cosets to map one group to a smaller, less detailed version is actually part of a larger pattern of mapping between one group to another, not-necessarily smaller group. Let’s give this process a name.

If we’re mapping between two groups (G,G)(G, *_G) and (Gˉ,Gˉ)(\bar{G}, *_{\bar{G}})6 with some function ff, we’d hope that ff preserves the “groupiness” of GG, namely that ff

  1. Respects the identity. So f(eG)=eGˉf(e_G) = e_{\bar{G}}.
  2. Respects inverses. So f(g1)=(f(g))1f(g^{-1}) = (f(g))^{-1} for all gGg \in G. In words, ff maps inverses in GG to inverses in Gˉ\bar{G}.
  3. Respects the operation. So f(aGb)=f(a)Gˉf(b)f(a *_G b) = f(a) *_{\bar{G}} f(b) for all a,bGa, b \in G.

It turns out we can derive properties 1 and 2 from 3, so we’ll restrict our attention to 3. Since the functions that satisfy property 3 preserve the “groupiness,” or structure that makes groups interesting, we’ll give them a name: group homomorphisms. Also, since the notation gets cumbersome, let’s drop the group-specific operations from property 3. Just keep in mind they are from different groups.

DEFINITION. A group homomorphism ff between groups GG and Gˉ\bar{G} is a mapping that preserves the group operation, i.e. f(ab)=f(a)f(b)f(ab) = f(a)f(b) for all a,bGa, b \in G

Now given this definition of a group homomorphism and what it does–preserve the operation–you might notice there’s a pretty natural way to understand the “identity-like” intuition from before. If f(b)=ef(b) = e (the identity of Gˉ\bar{G}) for some bGb \in G, then f(ab)=f(a)f(b)=f(a)e=f(a)f(ba)=f(b)f(a)=ef(a)=f(a). \begin{aligned} f(ab) &= f(a)f(b) = f(a)e = f(a) \\ f(ba) &= f(b)f(a) = ef(a) = f(a). \end{aligned} Here, bb seems to act like the identity under ff exactly because ff maps bb to the identity. As far as ff is concerned, it is the identity! Now zooming out, we’ll want to talk next about the set of all the elements in GG that map to the identity, which call the kernel of the group homomorphism ff, denoted kerf\ker f.

DEFINITION. The kernel of a group homomorphism ff are the set of elements of GG that ff maps to the identity in Gˉ\bar{G}, kerf={gG:f(g)=e}.\ker f = \{g \in G : f(g) = e\}.

<sidetrack> With this definition, allow me to detour a bit to tie this together with the intuition from the previous section. Consider the homomorphism from GG and its quotient group G/HG/H for some normal subgroup HH given by ϕ(g)=gH\phi(g) = gH. You might notice this is exactly the function that clumps elements into cosets now in symbols. You can verify it is indeed a homomorphism with the coset operation *.

Now observe that for every element hHh \in H that ϕ(h)=hH=H\phi(h) = hH = H, and HH is just the identity element of the quotient group. Conversely, if some element gGg \in G maps to the identity, ϕ(g)=H\phi(g) = H, then gH=HgH = H and gHg \in H necessarily. Thus inclusion goes both ways and kerϕ=H.\ker \phi = H. So, a normal subgroup HH that we hand-wavingly said before contained all the elements that are “identity-like” can now be formalized as exactly the elements that are the identity under ϕ\phi, exactly the kernel of ϕ\phi! In fact, we give this map ϕ\phi that formalizes our natural intuition a name: the natural homomorphism.

Furthermore, since this means every normal subgroup is the kernel of some homomorphism, namely the natural homomorphism, we might wonder if every kernel is a normal subgroup. And indeed it is.7 </sidetrack>

Now we’re ready for the entrée, the First Isomorphism Theorem. Let’s say we have a homomorphism ff from one group GG to another Gˉ\bar{G}. Consider the quotient group of cosets G/kerfG/\ker f (which we can create since kerf\ker f is normal): G/kerf={gkerf:gG}.G/\ker f = \{g\ker f : g \in G\}. Since kerf\ker f is the set of elements in GG that are the identity under ff, you should read each coset element gkerfg\ker f as “the set of elements in GG that are gg under ff.”

Given that gkerfg \ker f is all the elements that are gg under ff, they should behave exactly like gg under ff. That is, they should all map to where gg is mapped, namely f(g)f(g). You might visualize this with colors:

The First Isomorphism Theorem Visualized

Notice how G/kerfG/\ker f splits GG into clumps of elements that are all the same under ff. Accordingly, these clumps all map to the same element in Gˉ\bar{G}. Doesn’t it seem natural that there would be a one-to-one correspondence, and specifically in the context of groups, an isomorphism, between the clumps of colored circles on the left and where they map to colored squares on the right? That’s exactly the First Isomorphism Theorem! It states that,

THEOREM. Let GG and Gˉ\bar{G} be groups and f:GGˉf : G \rightarrow \bar{G} be a homomorphism between them. Then G/kerff(G),G / \ker f \cong f(G), where f(G)={f(g):gG}f(G) = \{f(g) : g \in G\} is the image of ff under GG. Explicitly, the isomorphism is given by the map gkerff(g)g\ker f \mapsto f(g).

Further, notice that because ff is a homomorphism, that all of Gˉ\bar{G} need not be hit, or in other words, we could have f(G)Gˉf(G) \neq \bar{G}. In our picture, there are potentially black squares that are not mapped to, but those don’t concern the First Isomorphism Theorem. All it says is that G/kerfG/\ker f is isomorphic onto the image (all the colored squares) of ff under GG, and that should feel natural!

For the sake of concreteness, let’s apply the First Isomorphism Theorem to an example. Consider the homomorphism f(x)=xmod4f(x) = x \bmod 4 from Z\mathbb Z to Z4\mathbb Z_4. Then kerf\ker f are all the multiplies of 44, since they are equivalent to 0mod40 \bmod 4. Now consider the quotient group Z/kerf=Z/4Z\mathbb Z/\ker f = \mathbb Z/4\mathbb Z. There are only four cosets in this quotient group, namely 0+4Z,1+4Z,2+4Z0 + 4\mathbb Z, 1+4\mathbb Z, 2+4\mathbb Z, and 3+4Z3 + 4\mathbb Z since it is cyclic. And of course, by the First Isomorphism Theorem, we know Z/4Zf(Z)=Z4\mathbb Z/4\mathbb Z \cong f(\mathbb Z) = \mathbb Z_4 when we map the coset 0+4Z0 + 4\mathbb Z to f(0)=0mod4=0Z4f(0) = 0 \bmod 4 = 0 \in \mathbb Z_4, the coset 1+4Z1 + 4\mathbb Z to f(1)=1mod4=1Z4f(1) = 1 \bmod 4 = 1 \in \mathbb Z_4, and so on. We’re just mapping the cosets to their representatives mod4\bmod 4, i.e. under the homomorphism.


Finally, in the spirit of milking this intuition for all it’s worth, we can also motivate ideals–what we get when we generalize the notion of normal subgroups over rings. Suppose we have a ring RR and we want to break it down into a smaller ring. We can first construct a smaller group by considering the additive abelian group (R,+)(R, +). Since it’s abelian, every subgroup II is normal and we know R/I={r+I:rR}R/I = \{r + I : r \in R\} is quotient group. Now, following intuition from earlier, II consists of elements that are identity-like, i.e. 00-like, in (R,+)(R, +). Of course, then we would expect that rIIrI \subseteq I and IrIIr \subseteq I for all rRr \in R, since left- and right-multiplication by something 00-like should yield something 00-like. This motivates the ideal, which absorbs multiplication.

And it turns out this ideal condition, that rI,IrIrI, Ir \subseteq I for all rRr \in R, is exactly necessary and sufficient to define multiplication on the quotient R/IR/I, such that we can create not just a quotient group, but also a quotient ring.

Final Thoughts

Hopefully some of these ideas and intuitions helped you digest what motivates quotient groups and normal subgroups, and how they can be applied and extended. They’ve helped me immensely in internalizing these concepts and how they generalize to other objects, like rings and varieties. Stay tuned for next time, where I’ll (most likely) talk about the rest of the isomorphism theorems. Oh yes–there’s more!

  1. In particular, HH is the identity of this “coset group,” and each coset a+Ha+H has the inverse a+H-a+H. The operation * also inherits associativity from ++

  2. I’ve written this in multiplicative notation since the group operation here, permutation composition, and more general operations have a multiplicative flavor. 

  3. HH is the identity and each coset aHaH has an inverse a1Ha^{-1}H where a1a^{-1} is the inverse permutation. The operation * inherits associativity from permutation composition. 

  4. Note that this does not mean gh=hggh = hg for some specific hHh \in H. It could be that gh1=h2ggh_1 = h_2g where h1h2h_1 \neq h_2 since we only require elements of HH to be identity-like, not exactly the identity. 

  5. For a proof, check out the first theorem of Arturo Magidin’s answer on Math.StackExchange. It’s incredibly thorough, formalizing the idea of yy-like with equivalence relations. 

  6. This just means GG is a group with the operation G*_G and Gˉ\bar{G} a group with the operation Gˉ*_{\bar{G}}. This way, we can be very explicit about which group operation occurs where. 

  7. First, for any homomorphism ff, the set kerf\ker f is a subgroup. By the subgroup test, if a,bkerfa, b \in \ker f, then f(ab1)=f(a)(f(b))1=ef(ab^{-1}) = f(a)(f(b))^{-1} = e so ab1kerfab^{-1} \in \ker f. To prove it is normal, it’ll suffice prove the logically-equivalent and easier-to-work-with condition gHg1=HgHg^{-1} = H I noted earlier, where H=kerfH = \ker f of some homomorphism ff. Then check out this Math.StackExchange post