A subgroup can fail to be normal in a spectacular way. Take the subgroup $H = \langle (1\ 2) \rangle$ of the symmetric group $S_3$, generated by the transposition that swaps $1$ and $2$. This is a perfectly legitimate subgroup of order $2$, but conjugating it by the permutation $\sigma = (1\ 2\ 3)$ gives

\begin{align*} \sigma H \sigma^{-1} &= \{e,\ \sigma (1\ 2) \sigma^{-1}\} = \{e,\ (2\ 3)\}, \end{align*}

which is a completely different subgroup of $S_3$. The group $S_3$ acts on its own subgroups by conjugation, and $H$ is not a fixed point of that action. This failure has consequences: you cannot form the quotient $S_3 / H$ in any sensible way, because the left and right cosets of $H$ do not agree, and so "cosets" do not multiply consistently. The concept of a normal subgroup exists precisely to identify which subgroups are immune to conjugation — the ones for which $gHg^{-1} = H$ for every $g \in G$ — and these are the subgroups that permit a well-defined quotient group.

[example: Failure of Left-Right Coset Agreement in $S_3$] To see concretely why non-normality obstructs the quotient construction, work out the cosets of $H = \{e, (1\ 2)\}$ in $S_3$. The left cosets are \begin{align*} eH &= \{e,\ (1\ 2)\}, \\ (1\ 2\ 3)H &= \{(1\ 2\ 3),\ (1\ 3)\}, \\ (1\ 3\ 2)H &= \{(1\ 3\ 2),\ (2\ 3)\}. \end{align*} The right cosets are \begin{align*} He &= \{e,\ (1\ 2)\}, \\ H(1\ 2\ 3) &= \{(1\ 2\ 3),\ (2\ 3)\}, \\ H(1\ 3\ 2) &= \{(1\ 3\ 2),\ (1\ 3)\}. \end{align*} The left coset $(1\ 2\ 3)H = \{(1\ 2\ 3),\ (1\ 3)\}$ and the right coset $H(1\ 2\ 3) = \{(1\ 2\ 3),\ (2\ 3)\}$ are not equal: the element $(1\ 3)$ appears in the left coset but not in the right. When left and right cosets differ, there is no natural group operation on the set of cosets: the "product" of two cosets would depend on which coset representative you choose, not just on the coset itself. [/example]

Normal subgroups are the correct level of generality for forming quotient groups, and they appear throughout algebra in roles that may not be immediately obvious — as kernels of homomorphisms, as building blocks in the composition series that classify groups up to extensions, and as the inputs to direct and semidirect product constructions.

## Definition

Before stating the definition, it is worth understanding what condition we actually need. We want to form a quotient $G/N$ whose elements are cosets and whose multiplication is

\begin{align*} (gN)(hN) &:= (gh)N. \end{align*}

For this to be well-defined, if we replace $g$ by $gn_1$ and $h$ by $hn_2$ for any $n_1, n_2 \in N$, the product must still land in the same coset:

\begin{align*} (gn_1)(hn_2) &= g(n_1 h)n_2 = g(h \cdot h^{-1} n_1 h) n_2 \in ghN \end{align*}

provided $h^{-1} n_1 h \in N$ for every $h \in G$ and every $n_1 \in N$. The condition $h^{-1}Nh \subset N$ for all $h \in G$ is exactly what we need — and combined with the reverse inclusion obtained by applying the condition to $h^{-1}$, this gives $h^{-1}Nh = N$. That is the definition.

[definition: Normal Subgroup] Let $G$ be a group and $N \le G$ a subgroup. We say $N$ is a **normal subgroup** of $G$, written $N \trianglelefteq G$, if \begin{align*} gNg^{-1} &= N \quad \text{for all } g \in G, \end{align*} where $gNg^{-1} := \{gng^{-1} : n \in N\}$ denotes the conjugate of $N$ by $g$. If additionally $N \ne G$, we say $N$ is a **proper normal subgroup**, written $N \triangleleft G$. [/definition]

Several equivalent characterizations of normality are worth recording immediately, because each reveals a different face of the condition.

[quotetheorem:787]

The equivalence of (i) and (ii) is slightly subtle: if $gNg^{-1} \subset N$ for all $g$, applying this to $g^{-1}$ gives $g^{-1}Ng \subset N$, i.e., $N \subset gNg^{-1}$, which together with the original inclusion forces equality. Condition (iii) is what makes the quotient construction work. Condition (v) gives an intrinsic, coordinate-free description: $N$ is normal in $G$ if and only if knowing that $n \in N$ forces the entire conjugacy class $\operatorname{Cl}(n) = \{gng^{-1} : g \in G\}$ to lie inside $N$.

[remark: Notation] We write $N \trianglelefteq G$ for normal subgroups and $N \triangleleft G$ for proper normal subgroups. The symbol $\le$ is reserved for arbitrary subgroups. An abelian group is its own center, so every subgroup of an abelian group is normal — condition (iii) is immediate, since $gN = Ng$ holds elementwise when $G$ is abelian. [/remark]

## The Quotient Group Construction

With the definition in hand, we can verify that normality is exactly the hypothesis needed to make the set of cosets into a group.

To appreciate why this theorem is nontrivial, consider what could go wrong. The set $G/N = \{gN : g \in G\}$ of left cosets is well-defined regardless of whether $N$ is normal. We want to define a binary operation by $(gN)(hN) := (gh)N$. The danger is that a coset has many representatives: $gN = g'N$ whenever $g' = gn$ for some $n \in N$. The product must not depend on which representatives we pick.

[quotetheorem:790]