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If G be an one empty set
(1) a(bc) = (ab)c for all a, b, c ∈ G
(2) For any a, b ∈ G, the equations ax = b and ya = b have solutions in G.
then
then
- a)<G, •> is not group
- b)<G. •> is a group
- c)<G,•> is a sub group
- d)<G, •> is not a sub group.
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
If G be an one empty set(1) a(bc) = (ab)c for all a, b, c ∈ G(2) ...
Let (1) and (2) hold. To show G is a group, we need prove existence of identity and
Let a ∈ G be any element inverse.
By (2) the equations ax = a
ya = a
have solutions in G.
Let x = e and y = f be the solutions.
ae = a
fa = a
Let now b e G be any element then again by (2) 
some x, y in G s.t.,
some x, y in G s.t.,ax = b
ya = b.
Now ax = b ⇒ f.(a.x) = f.b
⇒ (f.a).x = f.b
⇒ a.x = f.b
⇒ b = f.b
Again y.a = h
⇒ (y.a).e = b.e
⇒y.(a.e) = b.e
⇒ y.a = be
⇒ b = be
thus we have b = fb ... (i)
b = be ... (ii)
for any b ∈ G
Putting b = e in (i) and b = f in (ii) we get
e = fe
f = fe
⇒ e = f.
Hence ae = a = fa = ea
Hence ae = a = fa = ea
i.e., 
e ∈ G, s.t., ae = ea = a
e ∈ G, s.t., ae = ea = a⇒ e is identity.
Again, for any a ∈ G, and (the identity) e ∈ G, the equations ax = e and ya = e have solutions.
Let the solutions be x = a1, and y = a2
Again, for any a ∈ G, and (the identity) e ∈ G, the equations ax = e and ya = e have solutions.
Let the solutions be x = a1, and y = a2
then aa1 = e,
a2a = e
Now a1 = ea1 = (a2a)a1 = a2(aa1) = a2e = a2.
Hence aa1 = e = a1a for any a ∈ G
i.e., for any a ∈ G, 
some a1 ∈ G satisfying the above relations ⇒ a has an inverse. Thus each element has inverse and, by definition, G forms a group.
some a1 ∈ G satisfying the above relations ⇒ a has an inverse. Thus each element has inverse and, by definition, G forms a group.Most Upvoted Answer
If G be an one empty set(1) a(bc) = (ab)c for all a, b, c ∈ G(2) ...
Let (1) and (2) hold. To show G is a group, we need prove existence of identity and
Let a ∈ G be any element inverse.
By (2) the equations ax = a
ya = a
have solutions in G.
Let x = e and y = f be the solutions.
ae = a
fa = a
Let now b e G be any element then again by (2) some x, y in G s.t.,
some x, y in G s.t.,ax = b
ya = b.
Now ax = b ⇒ f.(a.x) = f.b
⇒ (f.a).x = f.b
⇒ a.x = f.b
⇒ b = f.b
Again y.a = h
⇒ (y.a).e = b.e
⇒y.(a.e) = b.e
⇒ y.a = be
⇒ b = be
thus we have b = fb ... (i)
b = be ... (ii)
for any b ∈ G
Putting b = e in (i) and b = f in (ii) we get
e = fe
f = fe
⇒ e = f.
Hence ae = a = fa = ea
Hence ae = a = fa = ea
i.e., e ∈ G, s.t., ae = ea = a
e ∈ G, s.t., ae = ea = a⇒ e is identity.
Again, for any a ∈ G, and (the identity) e ∈ G, the equations ax = e and ya = e have solutions.
Let the solutions be x = a1, and y = a2
Again, for any a ∈ G, and (the identity) e ∈ G, the equations ax = e and ya = e have solutions.
Let the solutions be x = a1, and y = a2
then aa1 = e,
a2a = e
Now a1 = ea1 = (a2a)a1 = a2(aa1) = a2e = a2.
Hence aa1 = e = a1a for any a ∈ G
i.e., for any a ∈ G, some a1 ∈ G satisfying the above relations ⇒ a has an inverse. Thus each element has inverse and, by definition, G forms a group.
some a1 ∈ G satisfying the above relations ⇒ a has an inverse. Thus each element has inverse and, by definition, G forms a group.Free Test
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Community Answer
If G be an one empty set(1) a(bc) = (ab)c for all a, b, c ∈ G(2) ...
Let (1) and (2) hold. To show G is a group, we need prove existence of identity and
Let a ∈ G be any element inverse.
By (2) the equations ax = a
ya = a
have solutions in G.
Let x = e and y = f be the solutions.
ae = a
fa = a
Let now b e G be any element then again by (2) some x, y in G s.t.,
some x, y in G s.t.,ax = b
ya = b.
Now ax = b ⇒ f.(a.x) = f.b
⇒ (f.a).x = f.b
⇒ a.x = f.b
⇒ b = f.b
Again y.a = h
⇒ (y.a).e = b.e
⇒y.(a.e) = b.e
⇒ y.a = be
⇒ b = be
thus we have b = fb ... (i)
b = be ... (ii)
for any b ∈ G
Putting b = e in (i) and b = f in (ii) we get
e = fe
f = fe
⇒ e = f.
Hence ae = a = fa = ea
Hence ae = a = fa = ea
i.e., e ∈ G, s.t., ae = ea = a
e ∈ G, s.t., ae = ea = a⇒ e is identity.
Again, for any a ∈ G, and (the identity) e ∈ G, the equations ax = e and ya = e have solutions.
Let the solutions be x = a1, and y = a2
Again, for any a ∈ G, and (the identity) e ∈ G, the equations ax = e and ya = e have solutions.
Let the solutions be x = a1, and y = a2
then aa1 = e,
a2a = e
Now a1 = ea1 = (a2a)a1 = a2(aa1) = a2e = a2.
Hence aa1 = e = a1a for any a ∈ G
i.e., for any a ∈ G, some a1 ∈ G satisfying the above relations ⇒ a has an inverse. Thus each element has inverse and, by definition, G forms a group.
some a1 ∈ G satisfying the above relations ⇒ a has an inverse. Thus each element has inverse and, by definition, G forms a group.
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If G be an one empty set(1) a(bc) = (ab)c for all a, b, c ∈ G(2) For any a, b ∈ G, the equations ax = b and ya = b have solutions in G.thena)<G, •> is not groupb)<G. •> is a groupc)<G,•> is a sub groupd)<G, •> is not a sub group.Correct answer is option 'B'. Can you explain this answer? for Mathematics 2025 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about If G be an one empty set(1) a(bc) = (ab)c for all a, b, c ∈ G(2) For any a, b ∈ G, the equations ax = b and ya = b have solutions in G.thena)<G, •> is not groupb)<G. •> is a groupc)<G,•> is a sub groupd)<G, •> is not a sub group.Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Mathematics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If G be an one empty set(1) a(bc) = (ab)c for all a, b, c ∈ G(2) For any a, b ∈ G, the equations ax = b and ya = b have solutions in G.thena)<G, •> is not groupb)<G. •> is a groupc)<G,•> is a sub groupd)<G, •> is not a sub group.Correct answer is option 'B'. Can you explain this answer?.
Solutions for If G be an one empty set(1) a(bc) = (ab)c for all a, b, c ∈ G(2) For any a, b ∈ G, the equations ax = b and ya = b have solutions in G.thena)<G, •> is not groupb)<G. •> is a groupc)<G,•> is a sub groupd)<G, •> is not a sub group.Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics.
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