Not all subgroups are created equal. Normal subgroups are the ones that play nicely with the rest of the group — and they’re the key to understanding when equations can be solved.

Normal subgroups

A subgroup $N$ of a group $G$ is normal (written $N \trianglelefteq G$) if it’s invariant under conjugation:

$$gNg^{-1} = N \quad \text{for all } g \in G$$

In other words, if you “rotate your perspective” by any element of the group, the subgroup looks the same.

Examples in $S_3$: - ${e, (123), (132)} = A_3$ is normal — it’s the alternating group (even permutations) - ${e, (12)}$ is not normal: $(123)(12)(132) = (23) \notin {e, (12)}$

All subgroups of abelian groups are normal (because $gng^{-1} = n$ when multiplication commutes).

Quotient groups

If $N \trianglelefteq G$, you can “collapse” $N$ to a point and get a new group $G/N$ — the quotient group. Its elements are the cosets $gN = {gn : n \in N}$.

Think of it as declaring all elements of $N$ to be “the same as the identity.” The remaining structure is the quotient.

Example: $S_3 / A_3 \cong \mathbb{Z}_2$. Collapsing the even permutations, you’re left with two classes: even and odd. The quotient group is just “even times even = even, odd times odd = even, even times odd = odd.”

The order of the quotient is $|G/N| = |G| / |N|$. This is Lagrange’s theorem in action.

Normal subgroups are highlighted. Click a node to see its elements and whether it’s normal.

Simple groups

A group is simple if it has no normal subgroups other than ${e}$ and itself. Simple groups are the atoms of group theory — every group can be built from simple groups (the Jordan-Holder theorem).

Key examples: - $\mathbb{Z}_p$ for any prime $p$ — simple (and cyclic) - $A_5$ (the alternating group on 5 elements, order 60) — simple but not cyclic

The distinction between “simple and cyclic” vs “simple but not cyclic” is exactly what determines solvability.

Why normality matters

In the Galois correspondence, each step down the chain of subgroups must be a normal subgroup of the one above it, and the quotient must be cyclic. This is because: - Normality ensures the quotient is well-defined (you can consistently “divide” by the subgroup) - Cyclicity corresponds to adjoining a single radical $\sqrt[n]{a}$

When these conditions hold all the way down, the equation is solvable by radicals. When they fail — when you hit a simple, non-cyclic group — you’re stuck.