The statement is asking to prove that a group G with the property that the square of every element is the identity (a^2 = e for all a in G) is an abelian group. An abelian group is a group in which the result of performing the group operation on any two group elements does not depend on their order (i.e., the group operation is commutative).

Here is a step-by-step proof:

1. Let a and b be any two elements in G. We want to show that ab = ba.

2. Since (ab)^2 = e, we have abab = e.

3. Multiplying both sides of abab = e on the right by a and b gives ababa = ea = a and ababb = eb = b.

4. But ababa = a implies abab = a^-1 = a (since a^2 = e), and ababb = b implies abab = b^-1 = b (since b^2 = e).

5. Therefore, we have ab = b^-1a^-1 = ba (since a^-1 = a and b^-1 = b), which shows that G is abelian.

Question

The statement is asking to prove that a group G with the property that the square of every element is the identity (a^2 = e for all a in G) is an abelian group. An abelian group is a group in which the result of performing the group operation on any two group elements does not depend on their order (i.e., the group operation is commutative).

Here is a step-by-step proof:

1. Let a and b be any two elements in G. We want to show that ab = ba.

2. Since (ab)^2 = e, we have abab = e.

3. Multiplying both sides of abab = e on the right by a and b gives ababa = ea = a and ababb = eb = b.

4. But ababa = a implies abab = a^-1 = a (since a^2 = e), and ababb = b implies abab = b^-1 = b (since b^2 = e).

5. Therefore, we have ab = b^-1a^-1 = ba (since a^-1 = a and b^-1 = b), which shows that G is abelian.

Knowee AI · Accepted Answer

The statement is asking to prove that a group G with the property that the square of every element is the identity (a^2 = e for all a in G) is an abelian group. An abelian group is a group in which the result of performing the group operation on any two group elements does not depend on their order (i.e., the group operation is commutative).

Here is a step-by-step proof:

1. Let a and b be any two elements in G. We want to show that ab = ba.

2. Since (ab)^2 = e, we have abab = e.

3. Multiplying both sides of abab = e on the right by a and b gives ababa = ea = a and ababb = eb = b.

4. But ababa = a implies abab = a^-1 = a (since a^2 = e), and ababb = b implies abab = b^-1 = b (since b^2 = e).

5. Therefore, we have ab = b^-1a^-1 = ba (since a^-1 = a and b^-1 = b), which shows that G is abelian.

8. Let 𝐺 be a group ith identity 𝑒 suoh that 𝑎2 = 𝑒 for all 𝑎 ∈ 𝐺, then 𝐺 is abelian.

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