Let G be any group. If N is a normal subgroup of G then prove that G/N is abelian iff del (G) is subset of N

Let \( G \) be any group, and \( N \) a normal subgroup of \( G \). We need to prove that:

\[ G/N \text{ is abelian if and only if } \delta(G) \subseteq N \]

where \( \delta(G) \) is the **derived subgroup** (also called the **commutator subgroup**) of \( G \), which is generated by all commutators \( [g, h] = g^{-1}h^{-1}gh \) for \( g, h \in G \).

### Proof:
#### (\( \Rightarrow \)): If \( G/N \) is abelian, then \( \delta(G) \subseteq N \)

Assume that \( G/N \) is abelian. This means that for all \( g, h \in G \), the cosets of \( g \) and \( h \) commute in the quotient group \( G/N \), i.e.,

\[
(gN)(hN) = (hN)(gN)
\]

Expanding both sides of this equation, we have:

\[
ghN = hgN
\]

which implies:

\[
ghN = hgN \implies g^{-1}h^{-1}gh \in N
\]

Notice that \( g^{-1}h^{-1}gh = [g, h] \) is the commutator of \( g \) and \( h \). Therefore, we have shown that for all \( g, h \in G \), the commutator \( [g, h] \in N \). Since the derived subgroup \( \delta(G) \) is generated by all such commutators, it follows that \( \delta(G) \subseteq N \).

#### (\( \Leftarrow \)): If \( \delta(G) \subseteq N \), then \( G/N \) is abelian

Now, assume that \( \delta(G) \subseteq N \). To prove that \( G/N \) is abelian, we need to show that for all \( g, h \in G \), the cosets \( gN \) and \( hN \) commute, i.e.,

\[
(gN)(hN) = (hN)(gN)
\]

We know that:

\[
(gN)(hN) = ghN \quad \text{and} \quad (hN)(gN) = hgN
\]

So, we need to show that \( ghN = hgN \). Equivalently, we want to prove:

\[
ghN = hgN \implies g^{-1}h^{-1}gh \in N
\]

But \( g^{-1}h^{-1}gh = [g, h] \) is the commutator of \( g \) and \( h \), and by assumption, \( \delta(G) \subseteq N \), meaning that all commutators of elements of \( G \) lie in \( N \). Therefore, \( [g, h] \in N \), and this implies that \( ghN = hgN \), which shows that the cosets \( gN \) and \( hN \) commute.

Thus, \( G/N \) is abelian.

### Conclusion:
We have shown that \( G/N \) is abelian if and only if \( \delta(G) \subseteq N \).

---

Would you like further details or have any questions about this proof?

Here are 5 related questions to consider:

1. What are some examples of groups where the derived subgroup \( \delta(G) \) is trivial?
2. Can you provide an example where \( G/N \) is abelian but \( G \) itself is non-abelian?
3. How does the derived series of a group relate to the commutator subgroup?
4. What are the properties of normal subgroups that allow for the quotient \( G/N \) to be well-defined?
5. How can you generalize this result to other types of group homomorphisms?

**Tip:** The derived subgroup \( \delta(G) \) helps measure how far a group is from being abelian, as its triviality implies that the group itself is abelian.

Let G be any group. If N is a normal subgroup of G then prove that G/N is abelian iff δ(G) is subset of N.

This detailed proof explains that G/N is abelian if and only if the derived subgroup δ(G) is a subset of N, where N is a normal subgroup of G. Key concepts include commutator subgroups and quotient groups, essential for understanding group theory at the undergraduate level.

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