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In a group G, a²b²=b²a² and a³b³=b³a³ hold for all a, b€G. Prove that the group is abelian
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In a group G, a²b²=b²a² and a³b³=b³a³ hold for all a, b€G. Prove that ...
Proof:
Let G be a group satisfying the conditions a²b² = b²a² and a³b³ = b³a³ for all a, b ∈ G.
Step 1: Show that (ab)² = a²b² for all a, b ∈ G.
Proof:
Consider the equation (ab)² = (ab)(ab).
Using the associative property, we can rewrite this as a(b(ab)).
Now, using the given condition a²b² = b²a², we can rewrite b(ab) as (ab)b.
So, we have a(b(ab)) = a((ab)b).
Using the associative property again, we can simplify this to (a(ab))b = (a²b²)b.
Finally, using the given condition a²b² = b²a², we can rewrite (a²b²)b as (b²a²)b.
Again, using the associative property, we get (b²(a²b))b = b²((a²b)b).
Since (a³b³ = b³a³) holds for all a, b ∈ G, we can rewrite (a²b)b as (ab)b².
So, we have b²((ab)b) = b²((ab)b²).
Using the associative property one last time, we can simplify this to b²(ab) = b²(a²b²).
Therefore, (ab)² = a²b².
Step 2: Show that (ab)³ = a³b³ for all a, b ∈ G.
Proof:
Consider the equation (ab)³ = (ab)(ab)(ab).
Using the associative property, we can rewrite this as a(b(ab))(ab).
Now, using the given condition a²b² = b²a², we can rewrite b(ab) as (ab)b.
So, we have a(b(ab))(ab) = a((ab)b)(ab).
Using the associative property again, we can simplify this to (a(ab))(b(ab)) = (a²b²)(b²a²).
Since (a³b³ = b³a³) holds for all a, b ∈ G, we can rewrite (a(ab)) as (ab)a².
Similarly, we can rewrite (b(ab)) as (ab)b².
So, we have (ab)a²(ab)b² = (ab)(a²b²)(b²a²).
Using the associative property one last time, we can simplify this to (ab)a²b²(ab)b² = (ab)(b²a²)(a²b²).
Therefore, (ab)³ = a³b³.
Step 3: Conclude that G is abelian.
Proof:
Using Step 1, we have (ab)² = a²b² for all a, b ∈ G.
Using Step 2, we have (ab)³ = a³b³ for all a, b ∈ G.
Now, consider the equation (ab)² = a²b².
By multiplying both sides by (ab), we get (ab)³ = a²b²(ab).
Using the given condition a²b² = b²a², we can rewrite a²b²(ab) as b²a²(ab).
Let G be a group satisfying the conditions a²b² = b²a² and a³b³ = b³a³ for all a, b ∈ G.
Step 1: Show that (ab)² = a²b² for all a, b ∈ G.
Proof:
Consider the equation (ab)² = (ab)(ab).
Using the associative property, we can rewrite this as a(b(ab)).
Now, using the given condition a²b² = b²a², we can rewrite b(ab) as (ab)b.
So, we have a(b(ab)) = a((ab)b).
Using the associative property again, we can simplify this to (a(ab))b = (a²b²)b.
Finally, using the given condition a²b² = b²a², we can rewrite (a²b²)b as (b²a²)b.
Again, using the associative property, we get (b²(a²b))b = b²((a²b)b).
Since (a³b³ = b³a³) holds for all a, b ∈ G, we can rewrite (a²b)b as (ab)b².
So, we have b²((ab)b) = b²((ab)b²).
Using the associative property one last time, we can simplify this to b²(ab) = b²(a²b²).
Therefore, (ab)² = a²b².
Step 2: Show that (ab)³ = a³b³ for all a, b ∈ G.
Proof:
Consider the equation (ab)³ = (ab)(ab)(ab).
Using the associative property, we can rewrite this as a(b(ab))(ab).
Now, using the given condition a²b² = b²a², we can rewrite b(ab) as (ab)b.
So, we have a(b(ab))(ab) = a((ab)b)(ab).
Using the associative property again, we can simplify this to (a(ab))(b(ab)) = (a²b²)(b²a²).
Since (a³b³ = b³a³) holds for all a, b ∈ G, we can rewrite (a(ab)) as (ab)a².
Similarly, we can rewrite (b(ab)) as (ab)b².
So, we have (ab)a²(ab)b² = (ab)(a²b²)(b²a²).
Using the associative property one last time, we can simplify this to (ab)a²b²(ab)b² = (ab)(b²a²)(a²b²).
Therefore, (ab)³ = a³b³.
Step 3: Conclude that G is abelian.
Proof:
Using Step 1, we have (ab)² = a²b² for all a, b ∈ G.
Using Step 2, we have (ab)³ = a³b³ for all a, b ∈ G.
Now, consider the equation (ab)² = a²b².
By multiplying both sides by (ab), we get (ab)³ = a²b²(ab).
Using the given condition a²b² = b²a², we can rewrite a²b²(ab) as b²a²(ab).
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In a group G, a²b²=b²a² and a³b³=b³a³ hold for all a, b€G. Prove that the group is abelian Related: Groups, Subgroups, Cyclic Groups and Permutation Groups - CSIR-NET Mathematical Sciences?
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