Understanding Groups with \(a^2 = e\) Property

The question asks about the nature of a group \(G\) where every element \(a\) satisfies the property \(a^2 = e\), where \(e\) is the identity element of the group. We need to determine if such a group is always a Ring, an Abelian Group, a Field, or a Non-abelian group.

Let's consider the given property: \(a^2 = e\) for all \(a \in G\). This means that for every element \(a\) in the group \(G\), multiplying the element by itself gives the identity element.

Recall the definition of an Abelian group. A group \(G\) is called an Abelian group (or commutative group) if the group operation is commutative, meaning that for any two elements \(a, b \in G\), \(ab = ba\).

Proving a Group with \(a^2 = e\) is Abelian

We are given that in group \(G\), \(a^2 = e\) for all \(a \in G\). Let's use this property to show that \(G\) is Abelian.

Consider any two arbitrary elements \(a, b \in G\).

Since \(a \in G\), the property tells us \(a^2 = a \cdot a = e\). Multiplying by \(a^{-1}\) on the left, we get \(a^{-1}(a \cdot a) = a^{-1}e\). By associativity and the identity property, this simplifies to \((a^{-1}a)a = a^{-1}\), so \(ea = a^{-1}\), which means \(a = a^{-1}\). Thus, for every element \(a\) in group \(G\), its inverse \(a^{-1}\) is equal to itself.

Similarly, for element \(b \in G\), we have \(b^2 = b \cdot b = e\), which implies \(b = b^{-1}\).

Now consider the product of the two elements \(a\) and \(b\), which is \(ab\). Since \(G\) is a group, the closure property ensures that \(ab\) is also an element of \(G\). Therefore, the element \((ab)\) must also satisfy the given property:

\((ab)^2 = e\)

Expanding the square, we get:

\((ab)(ab) = abab = e\)

We want to show that \(ab = ba\). Let's start with \(abab = e\) and manipulate the equation.

Multiply by \(a^{-1}\) on the left:

\(a^{-1}(abab) = a^{-1}e\)

\((a^{-1}a)bab = a^{-1}\)

\(ebab = a^{-1}\)

\(bab = a^{-1}\)

Since we know \(a^{-1} = a\), we can substitute this into the equation:

\(bab = a\)

Now, multiply by \(b^{-1}\) on the right:

\((bab)b^{-1} = ab^{-1}\)

\(ba(bb^{-1}) = ab^{-1}\)

\(bae = ab^{-1}\)

\(ba = ab^{-1}\)

Since we know \(b^{-1} = b\), we can substitute this into the equation:

\(ba = ab\)

We have successfully shown that for any two elements \(a, b \in G\), \(ab = ba\). This is the definition of an Abelian group.

Concluding the Group Type

Based on our proof, any group \(G\) where \(a^2 = e\) for all \(a \in G\) satisfies the commutative property \(ab = ba\) for all \(a, b \in G\). Therefore, such a group is an Abelian group.

Let's briefly consider the other options:

• Ring: A ring is a set with two binary operations (usually addition and multiplication) satisfying certain axioms. A group has only one operation. A group where \(a^2 = e\) is not necessarily a ring.
• Field: A field is a commutative ring where every non-zero element has a multiplicative inverse. A field is a more complex structure than just a group.
• Non-abelian group: A non-abelian group is one where the operation is not commutative, i.e., there exist \(a, b\) such that \(ab \neq ba\). Our proof showed that if \(a^2 = e\) for all \(a\), then \(ab = ba\) for all \(a, b\), so the group cannot be non-abelian.

Thus, the property \(a^2 = e\) for all \(a \in G\) uniquely identifies \(G\) as an Abelian group.