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If F is homomorphism of a group G into a group G' with kernal K, then
- a)K is a complex of G
- b)K is a subgroup of G but not a normal subgroup of G
- c)K is a normal subgroup of G.
- d)None of the above.
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If F is homomorphism of a group G into a group G with kernal K, thena)...
If F is a homomorphism of a group G into a group G' with kernel K, then the following properties hold:
- Kernel Definition:
The kernel of the homomorphism F, denoted as ker(F), is the set of all elements in G that are mapped to the identity element in G'.
In mathematical terms:
ker(F) = {g in G | F(g) = e'},
where e' is the identity element of G'. - Normal Subgroup:
The kernel K = ker(F) is a normal subgroup of G. This means that for all elements g, h in G:
gKg⁻¹ = K.
In other words, conjugating any element of the kernel by any element of G results in an element still within the kernel. - First Isomorphism Theorem:
The image of F, denoted by F(G), is isomorphic to the quotient group G / ker(F).
That is, G / ker(F) ≅ F(G).
This theorem provides a key relationship between the structure of G, its kernel, and its image under the homomorphism. - Injectivity:
The homomorphism F is injective (one-to-one) if and only if ker(F) = {e}, where e is the identity element of G.
This means that if the kernel is trivial (contains only the identity element), the homomorphism is injective. - Homomorphism Property:
For all elements a, b in G:
F(ab) = F(a)F(b).
This means the homomorphism preserves the group operation.
Conclusion:
- K is always a normal subgroup of G.
Hence option C is correct
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Community Answer
If F is homomorphism of a group G into a group G with kernal K, thena)...
Kernal of a Homomorphism
The kernal of a homomorphism plays a crucial role in understanding the properties of the homomorphism and the relationship between the two groups involved. Let's break down the options and explain why option 'C' is the correct answer.
a) K is a Complex of G
A complex is a set of elements in a group that satisfy a certain property. In this case, the kernal K is defined as the set of elements in G that map to the identity element in G under the homomorphism F. However, the kernal K does not necessarily form a complex of G. It is possible for K to be a complex, but it is not a guaranteed property.
b) K is a Subgroup of G but not a Normal Subgroup of G
A subgroup is a subset of a group that itself forms a group under the group operation. In this case, the kernal K is indeed a subgroup of G. To show this, we need to verify that K satisfies the three conditions to be a subgroup:
1. Closure: For any two elements a, b in K, their product ab must also be in K. This can be shown by considering the homomorphism F(a) and F(b), both of which are in G. Since G is a group, the product F(a)F(b) = F(ab) is also in G. Therefore, ab is in K.
2. Identity: The identity element of G, denoted as e, is always in K since F(e) = e', where e' is the identity element of G. Therefore, K contains the identity element.
3. Inverse: For any element a in K, its inverse a^(-1) must also be in K. This can be shown by considering F(a^(-1)), which is the inverse of F(a) in G. Since G is a group, the inverse element F(a^(-1)) is also in G. Therefore, a^(-1) is in K.
However, K may or may not be a normal subgroup of G. A normal subgroup is one that is invariant under conjugation by elements of G. In other words, for any element g in G and any element k in K, the conjugate gkg^(-1) is also in K. Whether K is normal or not depends on the specific homomorphism and the groups involved.
c) K is a Normal Subgroup of G
The correct answer is option 'C'. The kernal K is always a normal subgroup of G when considering a homomorphism F. To prove this, we need to show that for any element g in G and any element k in K, the conjugate gkg^(-1) is also in K.
Let's consider an arbitrary element g in G and an arbitrary element k in K. We know that F(g) and F(k) are in G since F is a homomorphism. Now, let's consider the homomorphism of the conjugate element:
F(gkg^(-1)) = F(g)F(k)F(g^(-1))
Since F(g) and F(k) are in G, their product F(g)F(k) is also in G. Similarly, the inverse F(g^(-1)) is also in G. Therefore, the product F(g)F(k)F(g^(-1)) is in
The kernal of a homomorphism plays a crucial role in understanding the properties of the homomorphism and the relationship between the two groups involved. Let's break down the options and explain why option 'C' is the correct answer.
a) K is a Complex of G
A complex is a set of elements in a group that satisfy a certain property. In this case, the kernal K is defined as the set of elements in G that map to the identity element in G under the homomorphism F. However, the kernal K does not necessarily form a complex of G. It is possible for K to be a complex, but it is not a guaranteed property.
b) K is a Subgroup of G but not a Normal Subgroup of G
A subgroup is a subset of a group that itself forms a group under the group operation. In this case, the kernal K is indeed a subgroup of G. To show this, we need to verify that K satisfies the three conditions to be a subgroup:
1. Closure: For any two elements a, b in K, their product ab must also be in K. This can be shown by considering the homomorphism F(a) and F(b), both of which are in G. Since G is a group, the product F(a)F(b) = F(ab) is also in G. Therefore, ab is in K.
2. Identity: The identity element of G, denoted as e, is always in K since F(e) = e', where e' is the identity element of G. Therefore, K contains the identity element.
3. Inverse: For any element a in K, its inverse a^(-1) must also be in K. This can be shown by considering F(a^(-1)), which is the inverse of F(a) in G. Since G is a group, the inverse element F(a^(-1)) is also in G. Therefore, a^(-1) is in K.
However, K may or may not be a normal subgroup of G. A normal subgroup is one that is invariant under conjugation by elements of G. In other words, for any element g in G and any element k in K, the conjugate gkg^(-1) is also in K. Whether K is normal or not depends on the specific homomorphism and the groups involved.
c) K is a Normal Subgroup of G
The correct answer is option 'C'. The kernal K is always a normal subgroup of G when considering a homomorphism F. To prove this, we need to show that for any element g in G and any element k in K, the conjugate gkg^(-1) is also in K.
Let's consider an arbitrary element g in G and an arbitrary element k in K. We know that F(g) and F(k) are in G since F is a homomorphism. Now, let's consider the homomorphism of the conjugate element:
F(gkg^(-1)) = F(g)F(k)F(g^(-1))
Since F(g) and F(k) are in G, their product F(g)F(k) is also in G. Similarly, the inverse F(g^(-1)) is also in G. Therefore, the product F(g)F(k)F(g^(-1)) is in
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If F is homomorphism of a group G into a group G with kernal K, thena)K is a complex of Gb)K isa subgroup of G but not a normal subgroup of Gc)K is a normal subgroup of G.d)None of the above.Correct answer is option 'C'. Can you explain this answer?
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