Abstract Algebra/Group Theory/Normal subgroups and Quotient groups

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In the preliminary chapter we discussed equivalence classes on sets. If the reader has not yet mastered this notion, he/she is advised to do so before starting this section.

Normal Subgroups[edit]

Recall the definition of kernel in the previous section. We will exhibit an interesting feature it possesses. Namely, let ak\,,\,k\in\ker\,\phi\leq G be in the coset a\ker\,\phi. Then there exists a k^\prime\in\ker\,\phi such that k^\prime a=ak for all a\in G. This is easy to see because a coset of the kernel includes all elements in G that are mapped to a paticular element. The kernel inspires us to look for what are called normal subgroups.


Definition 1: A subgroup N\leq G is called normal if gNg^{-1}=N for all g\in G. We may sometimes write N\trianglelefteq G to emphasize that N is normal in G.


Theorem 2: A subgroup N\leq G is normal if and only if gN=Ng for all g\in G.

Proof: By the definition, a subgroup is normal if and only if gNg^{-1}=N since conjugation is a bijection. The theorem follows by multiplying on the right by g.


We stated that the kernel is a normal subgroup in the introduction, so we had better well prove it!


Theorem 3: Let \phi\,:\,G\rightarrow G^\prime be any homomorphism. Then \ker\phi\leq G is normal.

Proof: Let k\in\ker\,\phi and g\in G. Then \phi(gkg^{-1})=\phi(g)e^\prime\phi(g)^{-1}=e^\prime, so gkg^{-1}\in\ker\phi, proving the theorem.


Theorem 4: Let G,G^\prime be groups and \phi\,:\, G\rightarrow G^\prime a group homomorphism. Then if H is a normal subgroup of \mathrm{im}\,\phi, then \phi^{-1}(H) is normal in G.

Proof: Let g\in G and h\in \phi^{-1}(H). Then \phi(ghg^{-1})=\phi(g)\phi(h)\phi(g)^{-1}\in H since H is normal in \mathrm{im}\,\phi, and so ghg^{-1}\in \phi^{-1}(H), proving the theorem.


Theorem 5: Let G,G^\prime be groups and \phi\,:\, G\rightarrow G^\prime a group homomorphism. Then if H is a normal subgroup of G, \phi(H) is normal in \mathrm{im}\,\phi.

Proof: Let g^\prime \in \mathrm{im}\,\phi And h\in H. Then if g\in G such that \phi(g)=g^\prime, we have g^\prime \phi(h) {g^\prime}^{-1}=\phi(g)\phi(h)\phi(g)^{-1}=\phi(ghg^{-1})=\phi(h^\prime)\in\phi(H) for some h^\prime\in H since H is normal. Thus g^\prime \phi(H){g^\prime}^{-1}=\phi(H) for all g^\prime\in\mathrm{im}\,\phi and so H is normal in \mathrm{im}\,\phi.


Corollary 6: Let G,G^\prime be groups and \phi\,:\, G\rightarrow G^\prime a surjective group homomorphism. Then if H is a normal subgroup of G, \phi(H) is normal in G^\prime.

Proof: Replace \mathrm{im}\,\phi with G^\prime in the proof of Theorem 5.


Remark 7: If H is a normal subgroup of G and K is a normal subgroup of H, it does not necessarily imply that K is a normal subgroup of G. The reader is invited to display a counterexample of this.


Theorem 8: Let G be a group and H,K be subgroups. Then

i) If H is normal, then HK=KH is a subgroup of G.
ii) If both H and K are normal, then HK=KH is a normal subgroup of G.
iii) If H and K are normal, then H\cap K is a normal subgroup of G.

Proof: i) Let H be normal. First, since for each k\in K, there exists h,h^\prime\in H such that kh=h^\prime k, so KH=HK. To show HK is a subgroup, let hk,h^\prime k^\prime\in HK. Then h^\prime k^\prime (hk)^{-1}=h^\prime k^\prime k^{-1} h^{-1}=h^\prime h^{\prime\prime}k^\prime k^{-1}\in HK for some h^{\prime\prime}\in H since H is normal, and so HK is a subgroup.

ii) Let g\in G and hk\in HK. Then since both H and K are normal, there exists h^\prime\in H\, ,\,k^\prime\in K such that ghkg^{-1}=ghg^{-1}k^\prime=gg^{-1}h^\prime k^\prime =h^\prime k^\prime\in HK. It follows that gHKg^{-1}=HK and so HK is normal.

iii) Let g\in G and h\in H\cap K. Then ghg^{-1}\in H since H is normal, and similarly ghg^{-1}\in K. Thus ghg^{-1}\in H\cap K and it follows that H\cap K is normal.


Examples of Normal Subgroups[edit]

In the following, let G be any group. Then G has associated with it the following normal subgroups.

i) The center of G, denoted Z(G), is the subgroup of elements which commute with all others. Z(G)=\{z\in G\mid (\forall g\in G) zg=gz\}. That Z(G) is a normal subgroup is easy to verify and is left to the reader.
ii) The commutator subgroup of G, denoted G^{(1)} or [G,G], is the subgroup generated by the subset \{[g,h]\mid g,h\in G\} where [g,h]=g^{-1}h^{-1}gh for all g,h\in G. For s\in G, we introduce the shorthand sgs^{-1}=g^s. Then we have s[g,h]s^{-1}=[g^s,h^s], such that for any product of commutators [g_1,h_1][g_2,h_2]...[g_n,h_n] where all elements are in G, we have s[g_1,h_1][g_2,h_2]...[g_n,h_n]s^{-1}=[g_1^s,h_1^s][g_2^s,h_2^s]...[g_n^s,h_n^s], and so G^{(1)} is normal.


Remark 9: We can iterate the commutator subgroup construction and define G^{(0)}=G and G^{(n)}=[G^{(n-1)},G^{(n-1)}] for all n\in \mathbb{N}. We will not use the commutator subgroup in future results in this book, so for us it is merely a curiosity.

Equivalence Relations on Groups[edit]

Why are normal subgroups important? In the preliminary chapter we discussed equialence relations and the associated set of equivalence classes. If G is a group and \sim is an equivalence relation, when does G/\sim admit a group structure? Of course we have to specify the multiplication on G/\sim. We will do so now.


Definition 10: Let G be a group and \sim is an equivalence relation on G, we define multiplication on the equivalence classes in G/\sim such that for all a,b\in G,

[a][b]=[ab]


This is indeed the only natural way to do it. Take the two equivalence classes, choose representatives, compute their product and take its equivalence class. The alert reader will have only one thing on his/her mind: is this well defined? For a general equivalence relation, the answer is no. The reader is invited to come up with an example. What is more interesting is when is it well defined? By the definition above, we obviously need the projection map \pi\,:\, G\rightarrow G/\sim defined by a\mapsto [a] to be a homomorphism. We can in fact condense the requirements down to two, both having to do with cancellation laws.


Theorem 11: Let G be a group and \sim an equivalence relation on G. Then G/\sim is a group under the natural multiplication if and only if for all a,b,g\in G

a\sim b\,\Leftrightarrow ag\sim bg \wedge ga\sim gb.

Proof: Assume G/\sim is a group. Since a\sim b \,\Leftrightarrow [a]=[b], the property follows from the cancellation laws in G. Assume now that the property holds. Then its multiplication rule is well defined, and must verify that G/\sim is a group. Let a,b,c\in G, then associativity is inherited from G:

([a][b])[c]=[ab][c]=[(ab)c]=[a(bc)]=[a][bc]=[a]([b][c]).

The identity in G/\sim is the equivalence class of e\in G, [e]:

[e][a]=[ea]=[a]=[ae]=[a][e].

Finally, the inverse of [a] is [a^{-1}]:

[g][g^{-1}]=[gg^{-1}]=[e]=[g^{-1}g]=[g^{-1}][g].

So G/\sim really defines a group structure, proving the theorem.


We will call an equivalence relation \sim compatible with G if G/\sim is a group. Then, G/\sim is called the quotient group of G by \sim. Also, as an immediate consequence, this makes \pi\,:\, G\rightarrow G/\sim into a homomorphism, but not just any homomorphism! It satisfies a universal property!


Commutative diagram showing the universal property satisfied by the projection homomorphism.

Theorem 12: Let \sim be en equivalence relation compatible with G, and \phi\,:\, G\rightarrow H a group homomorphism such that a\sim b \,\Rightarrow\, \phi(a)=\phi(b). Then there exists a unique homomorphism \tilde{\phi}\,:\, G/\sim\rightarrow H such that \phi=\tilde{\phi}\circ\pi.

Proof: In the preliminary chapter on set theory, we showed the corresponding statement for sets, so we know that \tilde{\phi} exists as a function between sets. We have to show that it is a homomorphism. This follows immediately: since \tilde{\phi}([a])=\phi(a) by commutativity, we have \tilde{\phi}([a])\tilde{\phi}([b])=\phi(a)\phi(b)=\phi(ab)=\tilde{\phi}([ab])=\tilde{\phi}([a][b]). As stated already, \tilde{\phi}([a])=\phi(a) shows uniqueness, proving the theorem.


Lemma 13: Let \sim be an equivalence relation on a group G such that a\sim b \,\Leftrightarrow ga=gb. Then H=[e] is a subgroup of G and a\sim b \,\Leftrightarrow\, a^{-1}b\in H\,\Leftrightarrow \, aH=bH.

Proof: First off, H is nonempty since e\in H. Let a,b\in H. Then b\sim e \,\Leftrightarrow\, e\sim b^{-1} by multiplying on the left by b^{-1}. Then since e\sim a we have a^{-1}\sim e by the same argument. Applying transitivity gives a^{-1}\sim b^{-1}. Finally, multiplying on the left by a gives e\sim ab^{-1}, giving ab^{-1}\in H and so H=[e] is a subgroup.

Assume a\sim b for a,b\in G. Then [a]=[b] implying [a^{-1}b]=[e]. Thus a^{-1}b\in [e]. Now assume a^{-1}b\in [e]. Then [a^{-1}b]=[e] and so [a]=[b] and finally a\sim b.

Assume a^{-1}b\in H. Then since H is a subgroup, we have a^{-1}bH=H and so aH=bH. Finally, assume aH=bH. Then a^{-1}bH=H. Since in paticular e\in H, this implies a^{-1}b\in H, completing the proof.


The mirror version using right cosets and the equivalence relation a\sim b \,\Leftrightarrow ag=bg and a\sim b \,\Leftrightarrow\, ab^{-1}\in H\,\Leftrightarrow \, Ha=Hb is completely analogous. Stating the theorem and writing out the proof is left to the reader as an exercise.


We have showed how an equialence relation defines a subgroup of G. In fact the equivalence classes are all cosets of this subgroup. We will now go the other way, and show how a subgroup defines an equivalence relation on G.


Lemma 14: Let H be a subgroup of a group G. Then,

i) a\sim_L b\,\Leftrightarrow\,a^{-1}b\in H\,\Leftrightarrow\,aH=bH is an equivalence relation such that a\sim_L b\,\Leftrightarrow\,ga\sim_L gb for all g\in G.
ii) a\sim_R b\,\Leftrightarrow\,ab^{-1}\in H\,\Leftrightarrow\,Ha=Hb is an equivalence relation such that a\sim_R b\,\Leftrightarrow\,ag\sim_R bg for all g\in G.

Proof: We will prove i). The proof for ii) is similar and is left as an exercise for the reader.

The fact that \sim is an equivalence relation and that a^{-1}b\in H\,\Leftrightarrow\,aH=bH was proven in the section on subgroups. Assume a\sim_L b. Then for all g\in G, (ga)^{-1}(gb)=a^{-1}g^{-1}gb=a^{-1}b\in H such that ga\sim_L gb. Now assume ga\sim_L gb, Then a^{-1}b=a^{-1}g^{-1}gb=(ga)^{-1}(gb)\in H such that a\sim_L b, completing the proof.


Theorem 15: For every equivalence relation \sim_L on G such that a\sim_L b\,\Leftrightarrow ga\sim_L gb, there exists a unique subgroup H of G such that G/\sim are precisely the left cosets of H.

Proof: This follows from Lemma 13 and Lemma 14.


Again, the mirror statement is completely analogous. Stating the theorem is left to the reader as an exercise.

Quotients with respect to Normal Subgroups[edit]

Lemma 16: An equivalence relation \sim on G is compatible with G if and only if [e]=H is a normal subgroup of G.

Proof: Assume \sim is compatible, g\in G and a\in H. Then a\sim e, and compatibility gives us gag^{-1}\sim gg^{-1}=e, and so gag^{-1}\in H. Since a is arbitrary, we obtain gHg^{-1}=H for all g\in G and so H is normal. Assume now that H is normal. Then aH=bH\,\Leftrightarrow a^{-1}b\in H, Ha=Hb\,\Leftrightarrow ab^{-1} \in H and aH=Ha for all a,b\in G. Using this, we obtain a\sim b\,\Leftrightarrow ab^{-1}=ab^{-1}bg(bg)^{-1}\sim e \,\Leftrightarrow ab^{-1}bg=ag\sim bg and similarily for the right hand case, so \sim is compatible with G.


Definition 17: When an equivalence relation is given by specifying a normal subgroup H, the quotient group with respect to this equivalence relation is denoted G/H. We then refer to G/H as the quotient of G with respect to H, or G modulo H. Note that this complies with previous definitions of this notation.

Multiplication in G/H is given as before as (aH)(bH)=(ab)H, with identity H and (aH)^{-1}=a^{-1}H for all a,b\in G.


Definition 18: Let H be a normal subgroup of G. Then we define the projection homomorphism \pi\,:\, G\rightarrow G/H by \pi(a)=aH for all a\in G.


Theorem 19: A subgroup is normal if and only if it is the kernel of some homomorphism.

Proof: We have already covered the left implication. For the right implication, assume H is normal. Then G/H is a group and we have the projection homomorphism \pi\,:\,G\rightarrow G/H as defined above. Since for all h\in H we have \pi(h)=hH=H, \ker\,\pi=H and so H is the kernel of a homomorphism.


Theorem 20: Let G,G^\prime be groups, \phi\,:\, G\rightarrow G^\prime a homomorphism and H a normal subgroup of G such that H\subseteq \ker\,\phi. Then there exists a unique homomorphism \tilde{\phi}\,:\,G/H\rightarrow G^\prime such that \tilde{\phi}\circ\pi=\phi.

Proof: This follows from Theorem 12 by letting a\sim b\,\Leftrightarrow aH=bH.

The Isomorphism Theorems[edit]

Commutative diagram showing the first isomorphism theorem. \tilde{\phi} is an isomorphism.

Theorem 21 (First Isomorphism Theorem): Let G,G^\prime be groups and \phi\,:\, G\rightarrow G^\prime a homomorphism. Then G/\ker\,\phi \approx \mathrm{im}\, \phi.

Proof: From Theorem 20 we have that there exists a unique homomorphism \tilde{\phi}\,:\, G/\ker\,\phi \rightarrow G^\prime such that \phi=\tilde{\phi}\circ\pi. We have to show that \tilde{\phi} is an isomorphism when we corestrict to \mathrm{im}\,\phi. This is immediate, since \phi(a)=\phi(b)\,\Leftrightarrow a\ker\phi=b\ker\phi by Lemma 13, so that \tilde{\phi} is injective, and for any g^\prime\in\mathrm{im}\,\phi there is a g\in G such that \phi(g)=\tilde{\phi}(g\ker\,\phi)=g^\prime so that it is surjective and therefore an isomorphism.


Lemma 22: Let G be a group, H a subgroup and K a normal subgroup of G. Then H\cap K is a normal subgroup of H.

Proof: Let h\in H and k\in H\cap K. Then hkh^{-1}\in H since h,k\in H and H is a subgroup and hkh^{-1}\in K since k\in K, h\in G and K is normal in G. Thus hkh^{-1}\in H\cap K and H\cap K is normal in H.


Theorem 23 (Second Isomorphism Theorem): Let G be a group, H a subgroup and K a normal subgroup of G. Then HK/K\approx \frac{H}{H\cap K}.

Proof: Define \phi\,:\, H\rightarrow HK/K by \phi(h)=hK for all h\in H. \phi is surjective since any element in HK can be written as hk with h\in H and k\in K, so \phi(h)=\pi(hk)=hkK=hK. We also have that \ker\,\phi=\{h\in H\mid hK=K\}=\{h\in H\mid h\in K\}=H\cap K and so HK/K\approx \frac{H}{H\cap K} by the first isomorphism theorem.


Lemma 24: Let G be a group, and let H,K be normal subgroups of G such that K\subseteq H\subseteq G. Then H/K is a normal subgroup of G/K.

Proof: Let h\in H and g\in G. Then ghg^{-1}=h^\prime for some h^\prime\in H since H is normal. Thus (gK)(hK)(gK)^{-1}=(ghg^{-1})K=h^\prime K, showing that H/K is normal in G/K.


Theorem 25 (Third Isomorphism Theorem) Let G be a group, and let H,K be normal subgroups of G such that K\subseteq H\subseteq G. Then \frac{G/K}{H/K}\approx G/H.

Proof: Let \phi\,:\, G/K\rightarrow G/H be given by \phi(gK)=gH. This is well defined and surjective since K\subseteq H, and is a homomorphism. Its kernel is given by \ker\,\phi=\{gK\in G/K\mid gH=H\}=\{gK\in G/K\mid g\in H\}=H/K, so by the first isomorphism theorem, \frac{G/K}{H/K}\approx G/H.


Theorem 26 (The Correspondence Theorem): Let G be a group and K be a normal subgroup. Now let A=\{H\leq G\mid K\leq H\} and B=\{H^\prime\mid H^\prime \leq G/K\}. Then \pi\,:\, H\mapsto \pi(H) is an order-preserving bijection from A to B.

Proof: We must show injectivity and surjectivity. For injectivity, note that if K\leq H, then \pi^{-1}(\pi(H))=HK=H, so if H_1,H_2\in A such that \pi(H_1)=\pi(H_2), then H_1=\pi^{-1}(\pi(H_1))=\pi^{-1}(\pi(H_2))=H_2, proving injectivity. For surjectivity, let H^\prime\in B. Then K\leq \pi^{-1}(H^\prime)\leq G, so that \pi^{-1}(H^\prime)\in A, and \pi(\pi^{-1}(H^\prime))=H^\prime, proving surjectivity. Lastly, since H_1\subseteq H_2 implies that \pi(H_1)\subseteq\pi(H_2), the bijection is order-preserving.


Note 27: The correspondence Theorem is sometimes called The Forth Isomorphism Theorem.


Theorem 28: Let H\in A from Theorem 26. Then H is normal if and only if \pi(H) is normal in G/K, and then G/H\approx \frac{G/K}{\pi(H)}.

Proof: Since \pi is surjective, H normal implies \pi(H) normal. Assume that \pi(H) is normal. Then \pi^{-1}(\pi(H))=H and so H is normal since it is the preimage of a normal subgroup. To show the isomorphism, let \phi\,:\,G\rightarrow \frac{G/K}{\pi(H)} be given by a composition of projections: \phi\,:\,\pi_{\pi(H)}\circ\pi_K. Then \ker\,\phi=\{g\in G\mid \phi(g)=\pi(H)\}=\{g\in G\mid \pi(g)\in \pi(H)\}=\pi^{-1}(\pi(H))=H, so by the first isomorphism teorem, G/H\approx\frac{G/K}{\pi(H)}.


Corollary 29: Let G be a group and H a normal subgroup. Then for any K^\prime\leq G/H there exists a unique subgroup K\leq G such that H\leq K\leq G and K=K^\prime H. Also, K is normal in G if and only if K^\prime is normal in G/H.

Proof: From Theorem 26 we have that the projection K\mapsto \pi(K)=K^\prime is a bijection, and since \pi(g)=gH for all g\in G, we have K=K^\prime H. The second part follows from Theorem 28.

Simple Groups[edit]

Definition 30: A group is called simple is it has no non-trivial proper normal subgroups.


Example 31: Every cyclic group \mathbb{Z}_p, where p is prime, is simple.


Definition 32: Let G be a group and H a normal subgroup. H is called a maximal normal subgroup if for any normal subgroup K of G, we have K\subseteq H.


Theorem 33: Let G be a group and H a normal subgroup. Then H is a maximal normal subgroup if and only if the quotient G/H is simple.

Proof: By Theorem 26 and Theorem 28, G/H has a nontrivial normal subgroup if and only if there exists a proper normal subgroup K of G such that H\leq K\leq G. That is, H is not maximal if ans only if G/H is not simple. The theorem follows.

Problems[edit]

Problem 1: Recall the unitary and special unitary groups from the section about subgroups. Define the projective unitary group of order n as the group PU(n)=U(n)/Z(U(n)). Similarily, define the projective special unitary group of order n as PSU(n)=SU(n)/Z(SU(n)).

i) Show that Z(SU(n))=SU(n)\cap Z(U(n)\approx \mathbb{Z}_n
ii) Using the second isomorphism theorem, show that PSU(n)\approx PU(n).