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Let G be a group of order 7 and φ(x) = x4, x ∈ G. Then f is
- a)not one-one
- b)not onto
- c)not a homomorphism
- d)one-one, onto and a homomorphism
Correct answer is option 'D'. Can you explain this answer?
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Let G be a group of order 7 and φ(x) = x4, x ∈ G. Then f isa)...
A group of prime order must be cyclic and every cyclic group is abelian. Then we can show that φ: G → G s.t. φ(x) = xn is an isomorphism if 0(G) and n and are co-prime.
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Let G be a group of order 7 and φ(x) = x4, x ∈ G. Then f isa)...
Let G be a group of order 7. Since the order of G is prime, G must be a cyclic group generated by one element, say g.
By Lagrange's theorem, the order of any subgroup of G must divide the order of G. Therefore, the possible orders of subgroups of G are 1 and 7.
If the order of a subgroup is 1, then the subgroup only contains the identity element. This subgroup is always a subgroup of any group. Therefore, G has a subgroup of order 1.
If the order of a subgroup is 7, then the subgroup is the entire group G itself. Therefore, G has a subgroup of order 7.
Thus, G has at least two subgroups: the trivial subgroup of order 1 and the entire group G of order 7.
By Lagrange's theorem, the order of any subgroup of G must divide the order of G. Therefore, the possible orders of subgroups of G are 1 and 7.
If the order of a subgroup is 1, then the subgroup only contains the identity element. This subgroup is always a subgroup of any group. Therefore, G has a subgroup of order 1.
If the order of a subgroup is 7, then the subgroup is the entire group G itself. Therefore, G has a subgroup of order 7.
Thus, G has at least two subgroups: the trivial subgroup of order 1 and the entire group G of order 7.
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Let G be a group of order 7 and φ(x) = x4, x ∈ G. Then f isa)...
But isme phi aur
f kya hai ye bhi to btana chahiy n
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Let G be a group of order 7 and φ(x) = x4, x ∈ G. Then f isa)not one-oneb)not ontoc)not a homomorphismd)one-one, onto and a homomorphismCorrect answer is option 'D'. Can you explain this answer?
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