Isomorphism between G/Z(G) and e

Introduction:

To determine whether G/Z(G) is isomorphic to e (the identity element), we need to understand the concepts of G/Z(G) and isomorphism.

G/Z(G):

G/Z(G) is the quotient group obtained by dividing the group G by its center Z(G). The center of a group G, denoted by Z(G), is the set of elements that commute with all other elements in G.

Isomorphism:

An isomorphism is a bijective map between two groups that preserves the group structure. In other words, if two groups G and H are isomorphic, there exists a one-to-one correspondence between their elements, and the group operations are preserved.

Proof:

To prove that G/Z(G) is isomorphic to e, we need to show that there exists a bijective map between G/Z(G) and e that preserves the group structure.

Step 1: Mapping to e

Since e is the identity element, every element in e is equal to e itself. Thus, we can define a map from G/Z(G) to e by assigning every element in G/Z(G) to e.

Step 2: Bijectivity

To prove bijectivity, we need to show that the map defined in Step 1 is both injective and surjective.

Injectivity:

Assume two elements x and y in G/Z(G) map to the same element in e. This implies that xZ(G) = yZ(G), which means xy^(-1) is in Z(G). Since Z(G) consists of elements that commute with all other elements in G, xy^(-1) must also commute with all elements in G. Therefore, xy^(-1) is in Z(G) if and only if xy^(-1) commutes with all elements in G. This implies that xy^(-1) = e, which further implies that x = y. Hence, the map is injective.

Surjectivity:

Since every element in G/Z(G) maps to e, the map is surjective.

Step 3: Preserving Group Structure

We need to show that the map defined in Step 1 preserves the group structure. Since every element in G/Z(G) maps to e, the group operation in G/Z(G) is preserved. Additionally, since e is the identity element, the identity element property is also preserved.

Conclusion:

Based on the proof provided above, we can conclude that G/Z(G) is isomorphic to e. The bijective map we defined satisfies the conditions of isomorphism, and it preserves the group structure.