Normal Subgroups

In the cosets lesson we saw that left and right cosets can differ. The subgroups for which they never differ are special: they are exactly the ones we can "divide by" to build a quotient group.

Definition. A subgroup $N$ of $G$ is normal, written $N \vartriangleleft G$ in many books, if $$gNg^{-1} = N \quad \text{for every } g \in G,$$ where $gNg^{-1} = \{gng^{-1} : n \in N\}$. Equivalent forms: $gN = Ng$ for every $g$; or, the test used in practice, $gng^{-1} \in N$ for every $g \in G$ and $n \in N$.

The element $gng^{-1}$ is a conjugate of $n$. Conjugating is "relabeling": in $S_{n}$, $\sigma\tau\sigma^{-1}$ is $\tau$ with every entry $i$ renamed $\sigma(i)$. So normal subgroups are the ones that look the same however you relabel the group.

Why it works

Three families of normal subgroups.
(1) Every subgroup of an abelian group: $gng^{-1} = ngg^{-1} = n$.
(2) Every subgroup of index 2. There are only two left cosets, $N$ and $G \setminus N$, and only two right cosets, also $N$ and $G \setminus N$. So for $g \in N$, $gN = N = Ng$, and for $g \notin N$, $gN = G \setminus N = Ng$.
(3) The center $Z(G) = \{z : zg = gz \text{ for all } g\}$, since $gzg^{-1} = z$. $\square$

In $D_{4}$, $N = \{e, r^{2}\}$: every left coset is also a right coset (the two panels match), so $N$ is normal. Here $r^{2}$ is the half turn, which commutes with everything.

Worked example: A normal and a non-normal subgroup of $D_{n}$

In $D_{n}$ ($n \ge 3$), is the rotation subgroup $\langle r \rangle$ normal? Is $\langle s \rangle = \{e, s\}$?

$\langle r \rangle$ has $n$ of the $2n$ elements, so index $2$: normal.
For $\langle s \rangle$ conjugate $s$ by $r$: $rsr^{-1} = r(sr^{-1}) = r(rs) = r^{2}s$, using $sr^{-1} = rs$.
$r^{2}s \ne s$ and $r^{2}s \ne e$ when $n \ge 3$, so $rsr^{-1} \notin \langle s \rangle$.

Answer: $\langle r \rangle$ is normal; $\langle s \rangle$ is not.

Worked example: $A_{n}$ is normal in $S_{n}$

Why is $A_{n}$ normal in $S_{n}$?

It has index $2$ ($|A_{n}| = n!/2$), so it is normal by the theorem.
Or directly: $\operatorname{sgn}(\sigma\tau\sigma^{-1}) = \operatorname{sgn}(\sigma)\operatorname{sgn}(\tau)\operatorname{sgn}(\sigma)^{-1} = \operatorname{sgn}(\tau)$, so conjugates of even permutations are even.

Answer: Index 2.

Worked example: Conjugating a cycle

In $S_{5}$, compute $\sigma\tau\sigma^{-1}$ for $\sigma = (1\;2\;3)$ and $\tau = (1\;4)(2\;5)$.

Rename each entry $i$ of $\tau$ as $\sigma(i)$: $1 \to 2$, $4 \to 4$, $2 \to 3$, $5 \to 5$.
$\sigma\tau\sigma^{-1} = (2\;4)(3\;5)$, the same cycle type as $\tau$.

Answer: $(2\;4)(3\;5)$.

Common mistake

Reading $gN = Ng$ as "$gn = ng$ for every $n$". Only the sets must match: $gn$ may equal a different element $n'g$. For instance $A_{3} = \{e, (1\;2\;3), (1\;3\;2)\}$ is normal in $S_{3}$, yet $(1\;2)(1\;2\;3) = (2\;3)$ while $(1\;2\;3)(1\;2) = (1\;3)$. A related trap: normality is not transitive. In $D_{4}$, $\{e, s\}$ is normal in $\{e, s, r^{2}, r^{2}s\}$ (index 2), which is normal in $D_{4}$ (index 2), but $\{e, s\}$ is not normal in $D_{4}$.

Key idea

$N$ is normal when $gNg^{-1} = N$ for all $g$, equivalently $gN = Ng$. Test with $gng^{-1} \in N$. Subgroups of abelian groups, subgroups of index 2, and centers are always normal.

Sources and further reading

Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), chapter 10, Normal Subgroups and Factor Groups (section 10.1)

MIT OpenCourseWare 18.703 Modern Algebra (Spring 2013), lecture on normal subgroups

John A. Beachy and William D. Blair, Abstract Algebra, online study guide (Northern Illinois University), section 3.8, Cosets, Normal Subgroups, and Factor Groups

Normal Subgroups

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