Group Homomorphisms and Normal Subgroup

Bunch homomorphisms and typical subgroups are essential ideas in unique variable-based math that assume a pivotal role in grasping the construction and conduct of gatherings. A gathering homomorphism is a plan between two gatherings that safeguards the gathering activity, giving a method for concentrating on the connections and similarities between gatherings.

Typical subgroups, then again, are subgroups that are invariant under formation by any component of the gathering, which takes into account the development of remainder gatherings and works on the investigation of gathering structure. Together, these ideas are fundamental for investigating bunch hypotheses' more profound perspectives, for example, remainder gatherings, factor gatherings, and the major hypothesis of homomorphism.

homomorphism

Definition of Group Homomorphism

A gathering homomorphism is a capability between two gatherings that safeguards the gathering activity. Officially, if (G,⋅) and (H,∗) are gatherings, a capability ϕ:G→H is known as a gathering homomorphism if, for all components a,b∈G, the accompanying condition holds:

ϕ(a⋅b)=ϕ(a)∗ϕ(b).

This implies that the picture of the result of two components under ϕ is equivalent to the result of their pictures. Bunch homomorphisms give a method for planning one gathering onto one more while keeping up with the logarithmic design of the gathering.

Examples of Group Homomorphisms

Here are a few examples of group homomorphisms:

• Character Homomorphism: For any gathering G, the capability ϕ:G→G characterized by ϕ(g)=g for all g∈G is a homomorphism. It maps each component to itself, safeguarding the gathering activity.

• Zero Guide: In the event that G is any gathering and H is an abelian bunch (or any gathering), the capability ϕ:G→H characterized by ϕ(g)=e H for all g∈G, where is the personality component in H, is a homomorphism. It maps each component of G to the personality component of H.

• Determinant Guide: Think about the overall direct gathering GL n (R) of invertible n×n frameworks over R and the multiplicative gathering of non-zero genuine numbers R ∗ . The determinant capability det:𝐺𝐿𝑛(𝑅→𝑅∗det:GL n (R)→R ∗ is a homomorphism, as det(𝐴𝐵)=det(𝐴)det(𝐵)det(AB)=det(A)det(B) for any frameworks An and B in 𝐺𝐿𝑛(𝑅)GLn(R).

• Projection Guide: Let G=Z×Z (the immediate result of two duplicates of the numbers) and H=Z. The projection map ϕ:Z×Z→Z characterized by ϕ((a,b))=a is a homomorphism since it protects the gathering activity (expansion for this situation), i.e., 𝜙((a1,𝑏1)+(𝑎2,𝑏2)) ==ϕ((a1+a2,b1+b 2))=a1+a2.

• Homomorphism from Numbers to Modulo Gatherings: Think about the gathering of whole numbers Z under expansion and the cyclic gathering Z/nZ (whole numbers modulo n). The capability ϕ:Z→Z/nZ characterized by 𝑛ϕ(k)=kmodn is a homomorphism, as ϕ(a+b)=(a+b)modn=(ϕ(a)+ϕ(b))modn.

Properties of Group Homomorphisms

Group homomorphisms have several important properties:

• Conservation of Character: A gathering homomorphism ϕ:G→H maps the personality component of G to the personality component of H. That is, assuming eG is the character in G and e H is the personality in H, then, at that point, ϕ(e G )=e H.

• Conservation of Inverses: A homomorphism jam inverses. In the event that ϕ:G→H is a homomorphism and g∈G, ϕ(g−1) =(ϕ(g)) −1 . This implies that taking the backwards in G and afterward applying ϕ yields a similar outcome as first applying ϕ and afterward taking the converse in H.

• Conservation of Gathering Activity: By definition, a homomorphism ϕ:G→H jam the gathering activity. For all a,b∈G,ϕ(a⋅b)=ϕ(a)∗ϕ(b), where ⋅ and ∗ mean the gathering activities in G and H, separately.

• Bit of a Homomorphism: The bit of a homomorphism ϕ:G→H is the arrangement of components in G that guide to the personality component in H. Officially, ker(ϕ)={g∈G∣ϕ(g)=e H }. The piece is a typical subgroup of G.

• Picture of a Homomorphism: The picture of a homomorphism ϕ:G→H is the arrangement of all components in H that are pictures of components in G. Officially, Im(ϕ)={ϕ(g)∣g∈G}. The picture is a subgroup of H.

Definition of Normal Subgroup

A typical subgroup N of a gathering G is a subgroup that is invariant under formation by any component of G. Officially, N is typical in G if for each component g∈G and each n∈N, the component gng - 1 is likewise in N. This property is meant as N◃G. Ordinary subgroups are significant in light of the fact that they consider the development of remainder gatherings, G/N, and are fundamental in concentrating on the construction and arrangement of gatherings.

Properties of Normal Subgroups

• Invariance under Formation: By definition, a subgroup N of G is ordinary if gNg −1 ⊆N for all g∈G. This implies gng −1 ∈N for each n∈N and g∈G.

• Convergence with Typical Subgroups: The crossing point of any assortment of ordinary subgroups of G is likewise a typical subgroup of G. If {Ni}i∈I is a group of typical subgroups of G, then ⋂i∈I Ni is ordinary in G.

• Subgroup of an Ordinary Subgroup: On the off chance that N is a typical subgroup of G and M is a subgroup of N, then M is likewise a typical subgroup of G.

• Remainder Gatherings: In the event that N is a typical subgroup of G, the remainder bunch G/N is clear cut. The arrangement of cosets G/N shapes a gathering with the activity characterized by (gN)⋅(hN)=(gh)N.

• Homomorphisms and Bits: The portion of a gathering homomorphism ϕ:G→H is a typical subgroup of G. On the off chance that 𝑁=ker(𝜙)N=ker(ϕ), ϕ(gng −1)=eH, guaranteeing gNg−1⊆N.
  

A mapping (function) f from a group (G,*) to a group (G',+) is called a group homomorphism (or group morphism) from G to G' if,  f(a*b) = f (a) +  f(b), for all a, b belong to set G.

Example on Group Homomorphisms and Normal Subgroup

1. The function f(x)=a^x from (R,+) to (R_o,*), where R is the set of all real numbers and R_o is the set of all real numbers excluding 0. as for any n,m ∈ R f(n+m)= a^n+m = aⁿ * a^m =f(n)*f(m), which is a homomorphism.
    
2. For cyclic group Z₃ = {0, 1, 2} and the Z(set of integers) with addition, 
      The function f(x)=x mod3 from Z₃ to (Z,+) is a group homomorphism.

NOTE - for a homomorphism f: G →G' 

• f is a monomorphism if f is injective (one-one).
• f  is Epimorphism, if f is surjective (onto).
• f is  Endomorphism if G = G'.
• G' is called the homomorphic image of the group G.

Solved Examples

Example 1:

Problem: Let φ: Z → Z6 be defined by φ(n) = [2n]6. Is φ a group homomorphism? If so, find ker(φ).

Solution:

To check if φ is a homomorphism, we need to verify φ(a + b) = φ(a) + φ(b) for all a, b ∈ Z.

φ(a + b) = [2(a + b)]6 = [2a + 2b]6 = [2a]6 + [2b]6 = φ(a) + φ(b)

So φ is a group homomorphism.

To find ker(φ), we solve φ(n) = [0]6

[2n]6 = [0]6

2n ≡ 0 (mod 6)

n ≡ 0 (mod 3)

Therefore, ker(φ) = 3Z

Example 2:

Problem: Prove that the center Z(G) of a group G is a normal subgroup of G.

Solution:

First, let's prove Z(G) is a subgroup:

e ∈ Z(G) as ex = xe for all x ∈ G

If a ∈ Z(G), then a^-1x = xa^-1 for all x ∈ G, so a^-1 ∈ Z(G)

If a, b ∈ Z(G), then (ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab) for all x ∈ G, so ab ∈ Z(G)

To prove normality, we need to show gZ(G)g^-1 ⊆ Z(G) for all g ∈ G:

Let z ∈ Z(G) and x ∈ G. Then:

(gzg^-1)x = g(zg^-1x) = g(g^-1xz) = xgzg^-1

So gzg^-1 ∈ Z(G)

Therefore, Z(G) is a normal subgroup of G.

Example 3:

Problem: Let G be a group and H be a subgroup of index 2 in G. Prove that H is normal in G.

Solution:

Since [G : H] = 2, G/H has only two cosets: H and aH for some a ∈ G \ H.

For any g ∈ G, gH must equal either H or aH.

If gH = H, then g ∈ H, so gHg^-1 = H.

If gH = aH, then g = ah for some h ∈ H. So:

gHg^-1 = (ah)H(ah)^-1 = aHa^-1 = aHa^-1

But aHa^-1 must be either H or aH. If it were aH, then H = a^-1(aH)a = Ha, implying a ∈ H, which contradicts our choice of a.

Therefore, gHg^-1 = H for all g ∈ G, so H is normal in G.

Example 4:

Problem: Let φ: G → H be a group homomorphism. Prove that if N is a normal subgroup of G containing ker(φ), then φ(N) is a normal subgroup of Im(φ).

Solution:

First, let's prove φ(N) is a subgroup of Im(φ):

eH ∈ φ(N) as eG ∈ N and φ(eG) = eH

If φ(n) ∈ φ(N), then φ(n)^-1 = φ(n^-1) ∈ φ(N) as n^-1 ∈ N

If φ(n1), φ(n2) ∈ φ(N), then φ(n1)φ(n2) = φ(n1n2) ∈ φ(N) as n1n2 ∈ N

To prove normality, we need to show φ(h)φ(N)φ(h)^-1 ⊆ φ(N) for all h ∈ Im(φ):

Let φ(g) = h for some g ∈ G and φ(n) ∈ φ(N).

φ(g)φ(n)φ(g)^-1 = φ(gng^-1)

Since N is normal in G, gng^-1 ∈ N, so φ(gng^-1) ∈ φ(N)

Therefore, φ(N) is a normal subgroup of Im(φ).

Example 5:

Problem: Let G be a finite group and p be the smallest prime dividing |G|. Prove that any subgroup H of G of index p is normal in G.

Solution:

Let [G : H] = p. By Lagrange's theorem, |G| = p|H|.

G acts on the set of left cosets G/H by left multiplication. This gives a homomorphism φ: G → Sp (symmetric group on p elements).

If ker(φ) = G, then φ is trivial and gH = H for all g ∈ G, so H is normal.

If ker(φ) ≠ G, then Im(φ) is a subgroup of Sp. By Lagrange's theorem, |Im(φ)| divides p!.

But |Im(φ)| = |G|/|ker(φ)| = p|H|/|ker(φ)|, which is divisible by p.

The only divisor of p! that's divisible by p is p itself (since p is the smallest prime dividing |G|).

Therefore, |Im(φ)| = p, so ker(φ) = H.

Since ker(φ) is always normal, H is normal in G.

Example 6:

Problem: Let G be an abelian group. Prove that every subgroup of G is normal.

Solution:

Let H be a subgroup of G.

To prove H is normal, we need to show gHg^-1 = H for all g ∈ G.

Let h ∈ H. We need to show ghg^-1 ∈ H for all g ∈ G.

Since G is abelian, gh = hg for all g, h ∈ G.

Therefore, ghg^-1 = hgg^-1 = h ∈ H.

Thus, gHg^-1 ⊆ H for all g ∈ G, so H is normal in G.

Example 7:

Problem: Let φ: G → H be a group homomorphism. If G is abelian, prove that Im(φ) is abelian.

Solution:

Let a, b ∈ Im(φ). Then there exist x, y ∈ G such that φ(x) = a and φ(y) = b.

We need to show ab = ba.

ab = φ(x)φ(y) = φ(xy) (since φ is a homomorphism)

ba = φ(y)φ(x) = φ(yx)

Since G is abelian, xy = yx

Therefore, φ(xy) = φ(yx), so ab = ba

Thus, Im(φ) is abelian.

Example 8:

Problem: Prove that the intersection of all conjugates of a subgroup H in G is a normal subgroup of G.

Solution:

Let N = ∩{gHg^-1 : g ∈ G} be the intersection of all conjugates of H.

First, let's prove N is a subgroup:

e ∈ N as e ∈ gHg^-1 for all g ∈ G

If a, b ∈ N, then a, b ∈ gHg^-1 for all g ∈ G, so ab ∈ gHg^-1 for all g ∈ G, thus ab ∈ N

If a ∈ N, then a ∈ gHg^-1 for all g ∈ G, so a^-1 ∈ gHg^-1 for all g ∈ G, thus a^-1 ∈ N

To prove normality, we need to show xNx^-1 ⊆ N for all x ∈ G:

Let n ∈ N and x ∈ G. Then for all g ∈ G:

xnx^-1 ∈ x(gHg^-1)x^-1 = (xg)H(xg)^-1

Since xg ∈ G, (xg)H(xg)^-1 is one of the conjugates in the intersection defining N.

Therefore, xnx^-1 ∈ N.

Thus, N is a normal subgroup of G.

Example 9:

Problem: Let G be a group of order pq, where p and q are distinct primes. Prove that G has a normal subgroup of order p or q.

Solution:

By Sylow's theorems, G has a Sylow p-subgroup P and a Sylow q-subgroup Q.

Let np be the number of Sylow p-subgroups and nq be the number of Sylow q-subgroups.

By Sylow's theorems:

np ≡ 1 (mod p) and np | q

nq ≡ 1 (mod q) and nq | p

The only possibility for np is either 1 or q.

The only possibility for nq is either 1 or p.

If np = 1, then P is the unique Sylow p-subgroup, which means it's normal in G.

If nq = 1, then Q is the unique Sylow q-subgroup, which means it's normal in G.

If np = q and nq = p, then |G| = pq = np + nq - 1, which is impossible.

Therefore, either P or Q must be normal in G.

Example 10:

Problem: Let G be a group and N be a normal subgroup of G. If G/N and N are both abelian, is G necessarily abelian? If not, provide a counterexample.

Solution:

G is not necessarily abelian. Here's a counterexample:

Let G be the dihedral group D8 of order 8, which is the symmetry group of a square.

G = ⟨r, s | r^4 = s^2 = e, srs = r^-1⟩

Let N = ⟨r^2⟩ = {e, r^2}, which is normal in G.

N is abelian (it's cyclic of order 2).

G/N has order 4 and is isomorphic to Klein four-group, which is abelian.

However, G itself is not abelian, as rs ≠ sr in D8.

Therefore, even if G/N and N are both abelian, G may not be abelian.  

Practice Problems - Group Homomorphisms and Normal Subgroup

1. Let G be a group and H be a subgroup of G. Prove that the intersection of all conjugates of H is a normal subgroup of G.

2. Let φ: Z → Z8 be defined by φ(n) = [3n]8. Determine if φ is a group homomorphism. If it is, find its kernel.

3. Prove that if G is an abelian group, then every subgroup of G is normal.

4. Let G be a group and Z(G) be its center. Prove that G/Z(G) is isomorphic to Inn(G) (the group of inner automorphisms of G).

5. Let φ: G → H be a group homomorphism. Prove that if N is a normal subgroup of G containing ker(φ), then φ(N) is a normal subgroup of Im(φ).

6. Let G be a group of order pq, where p and q are distinct primes. Prove that G has a normal subgroup of order p or q.

7. Let G be a group and H be a subgroup of index 2 in G. Prove that H is normal in G.

8. Let φ: G → H be a group homomorphism. If G is abelian, prove that Im(φ) is abelian.

9. Let G be a group and let H be a subgroup of G. Show that H is normal if and only if gH = Hg for all g ∈ G.

10. Let G be a group and let H be a subgroup of G. Show that if H is of index 2 in G, then H is normal in G.

Conclusion

Group homomorphisms and normal subgroups are fundamental concepts in group theory. Homomorphisms preserve the group structure, while normal subgroups are special types of subgroups that are invariant under conjugation. The relationship between these two concepts is essential for understanding the structure of groups and their quotient groups.

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