Normal Subgroups and Group Quotients

[ This corresponds to approximately chapter V of the old blog. ]

We’ve already seen that if H ≤ G is a subgroup, then G is a disjoint union of (left) cosets of H in G. We’d like to use the set of these cosets to form a group. Naturally, we define:

xHyH := (xy)H.

The only problem which occurs is whether this is well-defined (note: this will be a recurring theme in algebra). Explicitly, if xHx’H and yHy’H, then is it true that (xy)H = (x’y’)H? We won’t repeat the argument here, but the end result is that the definition makes sense if and only if H is normal in G:

Definition. A subgroup H of G is said to be a normal subgroup  (written $H \triangleleft G$) if for any $g\in G$ and $h\in H$, we have $ghg^{-1} \in H$.

If $N \triangleleft G$, then the set of cosets G/N can be made into a group via (gN)*(g’N) := (gg’)N.

Equivalently, we can say $gHg^{-1} \subseteq H$ for any g in G. In fact, this implies $gHg^{-1} = H$ for any g in G. [ To see the reverse inclusion, replace g by its inverse to obtain $g^{-1}Hg \subseteq H \implies H \subseteq gHg^{-1}$. ] So in fact, saying H is a normal subgroup is equivalent to saying the left cosets gH are identical to the right cosets Hg.

Be careful: for general subgroup H of G, it’s possible to find an element g such that $gHg^{-1} \subset H$ strictly. E.g. consider $G = GL_2(\mathbf{R})$ and the subgroup $H = \left\{\begin{pmatrix} 1 & n \\ 0 & 1\end{pmatrix} : n \in \mathbf{Z}\right\}$. The element $g = \begin{pmatrix} 1 & 0 \\ 0 & 1/2\end{pmatrix}$ then gives $gHg^{-1} = \left\{\begin{pmatrix} 1 & 2n \\ 0 & 1\end{pmatrix} : n\in \mathbf{Z}\right\}$ which is properly contained in H. If you think this contradicts the previous paragraph, mull over it a bit more.

Examples

Let’s consider the examples in the previous posts. In the abelian case, all subgroups are normal since $ghg^{-1} = gg^{-1}h = h$. But let’s see what their quotient groups look like.

1. For mZZ, where m>0, the quotient is precisely Z/m. In fact, this is precisely the definition of Z/m.
2. For Q>0Q*, it’s quite easy to prove that $\mathbf{Q}^* \cong \mathbf{Q}^{>0} \times {\pm 1}$. So the quotient is (isomorphic to) Z/2. For Q*2 < Q>0, the resulting quotient is a product of infinitely many copies of Z/2: one for each prime p.
3. The quotient of a cyclic group is still cyclic, so for k·(Z/m) < Z/m (where k|m) the resulting quotient is Z/k
4. For (Z/21)*2 < (Z/21)*, the resulting group quotient has order 4. Since the square of every element lies in the subgroup, every element of the quotient has order 2. By the previous post, we see that the group is Z/2 × Z/2.

For the non-abelian cases:

1. The image of Sm in Sn is not normal if 1 < mn. This is because although (1,2) lies in the subgroup, $(2,n)(1,2)(2,n)^{-1} = (1,n)$ does not.
2. The subgroup VS4, where V = {e, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3)}, is a normal subgroup. To see why, our first post on permutations mentioned that if g and h are permutations, then $ghg^{-1}$ has the same cycle structure as h. But V already contains all permutations with cycle structure 2+2. The quotient has order 6. Check that it’s isomorphic to S3. [ Worst come to worst, just draw the 6 × 6 multiplication table. ]
3. For SLn(R) < GLn(R), the subgroup is normal since if det(h)=1, then $\det(ghg^{-1}) = \det(g)\det(h)\det(g)^{-1} = \det(g)\det(g)^{-1} = 1$.

Properties of Normal Subgroups

Some basic properties include:

1. The intersection of normal subgroups of G is also normal.
2. If N is a normal subgroup of G, and H is an arbitrary subgroup of G, then N ∩ H is normal in H.
3. Normality is not transitive, i.e. we can find H normal in G, N normal in H, but N is not normal in G.

The proofs of (1) & (2) are easy and left as exercises. For (3), pick $\{e, (1,2)\} \triangleleft V \triangleleft S_4$. But clearly {e, (1,2)} is not normal in S4. Property (1) has an interesting implication: if S is some subset of G, we let N(S) be the intersection of all normal subgroups of G containing S. As beforeN(S) is called the normal subgroup of G generated by S and is the “smallest” normal subgroup of G containing S. This will come in handy later.

Symbolic Visualisation

I find the following visualisation helpful. For normal subgroup $N \triangleleft G$, since gNg-1N, we have gNNg and thus:

$(gN)(g'N) = g(Ng')N = g(g'N)N = (gg')(NN) \subseteq (gg')N$

where for any subsets ST of G, we define $ST = \{xy : x\in S, y\in T\}$. The inclusion $NN \subseteq N$ is equivalent to saying N is closed under group product *. Actually we even have NNN, but we don’t need that to show that * is well-defined on the set of cosets.

The above symbolic visualisation will be helpful in the following.

Proposition. Let H, K ≤ G be subgroups. In general, HK is not a subgroup of G. However:

1. if H (or K) is normal in G, then HK is a subgroup;
2. if H and K are both normal in G, then HK is normal in G.

Sketch of Proof.

To show that HK is not a subgroup in general, let H = {e, (1,2)} and K = {e, (2,3)} in S3. Now, for (1), if H is normal in G, then for any elements kk’ of K, we have:

$(Hk)(Hk') = H(kH)k' = H(Hk)k' \subseteq H(kk') \subseteq HK$.

For inverse, $(Hk)^{-1} = k^{-1}H = Hk^{-1} \subseteq HK$.

To prove (2), if H and K are both normal, then for any element g of G, we have:

$g(HK) = (gH)K = (Hg)K = H(gK) =H(Kg) = (HK)g$

which completes the proof rather neatly and you can easily recall it any time. ♦

Group Products

Consider normal subgroups $H, K\triangleleft G$ as above. If we assume they intersect trivially ($H\cap K = \{e\}$) then elements of H must commute with elements of K since:

$(hk)(kh)^{-1} = hkh^{-1}k^{-1} = \overbrace{(hkh^{-1})}^{\in K}k^{-1} = h\overbrace{(kh^{-1}k^{-1})}^{\in H}$

must lie in both H and K and so must be the identity. Furthermore, the multiplication map fH × K → HK is an isomorphism. Why? Because

• f maps (hk) * (h’k’) = (hh’, kk’) to hh’kk’ = hkh’k’, which is the product of f(hk) and f(h’k’). [ A more proper notation would be f((hk)) but you know what we mean here. ]
• The map is surjective by definition of HK.
• The map is injective since $hk = h'k' \implies h'^{-1}h = k'k^{-1} \in H\cap K$. So it must be the identity and h=h’k=k’.

Conclusion. If H, K are normal subgroups of G with trivial intersection, then HK is isomorphic to H × K via the multiplication map (h, k) → hk.

Conversely, if you have a group product G = H × K, then clearly H × {e} and {e} × K are two normal subgroups of G which intersect trivially. Thus we have found a criterion for identifying group products: given G, can we express it as a product of two groups?

Now let’s loosen the condition a little and still say $H \cap K = \{e\}$, but now H is normal in G though K may not be. What happens to the multiplication map fH × K → HK? It’s still surjective by definition. Injectivity follows since H and K intersect trivially. The only problem is that f may not respect the group product on both sides since elements of H don’t commute with elements of K now (the earlier reasoning falls apart).

So we have a bijective map between two groups which doesn’t respect the algebra. But all is not lost, for it turns out we’ll get what’s called the semi-direct product. This will be covered in a later post.

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