Let G be a group. An element of G that can be expressed in the form aba−1b−1 for some a,b∈G is a commutator in G. The preceding exercise shows that there is a smallest normal subgroup C of a group G containing all commutators in G; the subgroup C is the commutator subgroup of G. Show that G/C is an abelian group.

To show that \( G / C \) is an abelian group, we need to prove that for any two elements \( g_1C \) and \( g_2C \) in the quotient group \( G / C \), their product satisfies the commutative property:

\[
g_1C \cdot g_2C = g_2C \cdot g_1C
\]

### Step-by-step Proof:

1. **Recall the definition of \( G / C \)**:
   - \( C \) is the commutator subgroup of \( G \). It is the smallest normal subgroup of \( G \) that contains all commutators of \( G \). A commutator is an element of the form \( aba^{-1}b^{-1} \) for some \( a, b \in G \).
   - The quotient group \( G / C \) consists of cosets of the form \( gC \), where \( g \in G \).

2. **Commutativity in \( G / C \)**:
   - We want to show that for any \( g_1, g_2 \in G \), the cosets \( g_1C \) and \( g_2C \) commute, i.e., \( g_1C \cdot g_2C = g_2C \cdot g_1C \).

3. **Expand both products**:
   \[
   g_1C \cdot g_2C = (g_1g_2)C
   \]
   and
   \[
   g_2C \cdot g_1C = (g_2g_1)C.
   \]

4. **Relating these two products**:
   - The products \( g_1g_2 \) and \( g_2g_1 \) differ by the commutator of \( g_1 \) and \( g_2 \):
     \[
     g_1g_2 = g_2g_1 \cdot [g_1, g_2],
     \]
     where \( [g_1, g_2] = g_1g_2g_1^{-1}g_2^{-1} \) is the commutator of \( g_1 \) and \( g_2 \).

5. **Use the fact that \( C \) contains all commutators**:
   - Since \( C \) is the commutator subgroup, the commutator \( [g_1, g_2] \in C \). Therefore, we can write:
     \[
     g_1g_2 = g_2g_1 \cdot c
     \]
     for some \( c \in C \).

6. **In the quotient group \( G / C \)**:
   - In the quotient group, elements of \( C \) are equivalent to the identity element. Therefore, we have:
     \[
     (g_1g_2)C = (g_2g_1c)C = (g_2g_1)C,
     \]
     because \( c \in C \), so multiplying by \( c \) does not change the coset.

7. **Conclusion**:
   - Thus, \( g_1C \cdot g_2C = g_2C \cdot g_1C \), which means that \( G / C \) is an abelian group.

### Final Statement:
Since \( G / C \) satisfies the commutative property, we conclude that \( G / C \) is an abelian group.

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Would you like a more detailed explanation of any specific part of this proof?

Here are 5 related questions to expand on this topic:
1. What is the significance of a commutator in group theory?
2. Why is the commutator subgroup the smallest normal subgroup containing all commutators?
3. How can we interpret the structure of the quotient group \( G / C \) geometrically?
4. How does the commutator subgroup relate to solvable groups?
5. Can we provide an example of a group \( G \) where the commutator subgroup is non-trivial?

**Tip**: A group is abelian if and only if its commutator subgroup is trivial.

Let G be a group. An element of G that can be expressed in the form aba−1b−1 for some a, b ∈ G is a commutator in G. The preceding exercise shows that there is a smallest normal subgroup C of a group G containing all commutators in G; the subgroup C is the commutator subgroup of G. Show that G/C is an abelian group.

Explore how to prove that the quotient group G/C, where C is the commutator subgroup, is abelian. This proof involves key concepts from group theory, such as commutators and quotient groups. Suitable for undergraduate students studying abstract algebra.

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