Some subgroups sit inside their group so neatly that you can't knock them out of place, no matter how you "look at them" from the rest of the group. Those are the normal subgroups, and they are the key that unlocks one of the most important constructions in all of algebra. A note on conventions before we start: as on Symmetric Groups, products of permutations are composed right to left, so in $\sigma\tau$ we apply $\tau$ first and then $\sigma$.
Let $G$ be a group. Here's the definition everything on this page revolves around.
The symbol for "normal subgroup" is a triangle eating an equals sign. $\unlhd$. Nom.
A subgroup $H \le G$ is normal if, $\forall g \in G$, $gHg^{-1} = \{ghg^{-1} : h \in H\} \subseteq H$.
Notation: $H \unlhd G$.
The expressions in that definition have names of their own.
$ghg^{-1}$ is the conjugate of $h$ by $g$, and $gHg^{-1}$ is the conjugate of $H$ by $g$.
So $H$ is normal when conjugating any element of $H$, by any element of the whole group, always lands you back in $H$. Why should we care?
Normal subgroups are precisely what are needed to define "quotient groups" $G/H$, which are analogues of $\mathbb{Z}/n\mathbb{Z}$ for other groups.
Let's start with the two subgroups every group has.
Suppose $G$ is any group and let $H = \{e\}$. Is $H$ normal?
Then $\forall g \in G$, $geg^{-1} = gg^{-1} = e \in H$. Therefore $\{e\} \unlhd G$.
Suppose $G$ is any group and let $H = G$. Is $H$ normal?
Then $\forall g \in G$, $h \in H$, $ghg^{-1} \in G = H$. Therefore $G \unlhd G$.
Some groups have no other normal subgroups at all. They get a special name.
$A_5$ has $60$ elements, and mathematicians looked at it and said "yep, that one's simple."
Groups $G$ whose only normal subgroups are $\{e\}$ and $G$ are called simple groups.
Some examples of simple groups are:
Simple groups are the "atoms" that every finite group is built from, which is why mathematicians worked so hard to find them all. For now, back to examples. In an Abelian group, the conjugate collapses immediately.
Show that if $G$ is Abelian then every subgroup $H \le G$ is normal.
$\forall h \in H, g \in G$, $ghg^{-1} = (gg^{-1})h = h \in H$ (since $G$ is Abelian). Therefore $gHg^{-1} \subseteq H$.
So the interesting cases live in non-Abelian groups. The smallest one is $S_3$.
$S_3$ is the smallest non-Abelian group there is, and it's already refusing to cooperate.
$G = S_3 = \{e, (12), (13), (23), (123), (132)\}$ and $H = \langle (123) \rangle = \{e, (123), (132)\}$. Is $H$ normal?
Note: $H = A_3$ (the subgroup of $S_3$ consisting of all even permutations).
$\forall g \in G, h \in H$, $h$ is even, so $ghg^{-1}$ is even $\Rightarrow ghg^{-1} \in H$. (Conjugating doesn't change parity: $g$ and $g^{-1}$ have the same parity, so their contributions cancel.)
Conclusion: $H \unlhd G$.
$G = S_3$ and $H = \langle (12) \rangle = \{e, (12)\}$. Is $H$ normal?
Try conjugating $(12)$ by $(23)$. Composing right to left, $(23)(12) = (132)$ and then $(132)(23) = (13)$, so
$$(23)(12)(23)^{-1} = (23)(12)(23) = (132)(23) = (13) \notin H.$$
Conclusion: $H \not\unlhd G$.
One element of $G$ knocking one element of $H$ outside is all it takes to break normality. See it happen below: each permutation is drawn as a layer of strands, and $ghg^{-1}$ stacks three layers.
Pick a subgroup $H$, an element $g$, and an element $h \in H$. Follow the strands top to bottom to compute $ghg^{-1}$: reading right to left, $g^{-1}$ acts first.
Is $\langle (13) \rangle = \{e, (13)\}$ normal in $S_3$? If not, find a witness: an element $g$ with $g(13)g^{-1} \notin \langle (13) \rangle$.
Not normal. Take $g = (12)$. Composing right to left, $(12)(13) = (132)$ and $(132)(12) = (23)$, so
$$(12)(13)(12)^{-1} = (12)(13)(12) = (23) \notin \langle (13) \rangle.$$
The same thing happens for $\langle (23) \rangle$: none of the three subgroups of order $2$ in $S_3$ is normal.
The definition only asks for $gHg^{-1} \subseteq H$. It turns out that's secretly much stronger than it looks. First, some notation for multiplying a whole set by an element.
$\forall g, g' \in G$, $\forall S \subseteq G$ define:
With that notation, here are three ways of saying the same thing.
Suppose $G$ is a group and $H$ is a subgroup of $G$. The following statements are equivalent:
Statement (ii) is the surprise: if conjugating never takes you outside $H$, then it actually hits all of $H$. To prove a chain of equivalences, we show (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i).
Proving $gHg^{-1} = H$ is a job for double inclusion: we already have $gHg^{-1} \subseteq H$, so we only need $H \subseteq gHg^{-1}$.
Proof.
(i) $\Rightarrow$ (ii): Suppose (i) holds. We need to show that $\forall g \in G$, $H \subseteq gHg^{-1}$. Note that $\forall g \in G$, we have $g^{-1} \in G$, so $g^{-1}H(g^{-1})^{-1} = g^{-1}Hg \subseteq H$. Therefore, $\forall h \in H$, $g^{-1}hg \in H \Rightarrow g^{-1}hg = h'$ for some $h' \in H \Rightarrow h = gh'g^{-1} \Rightarrow h \in gHg^{-1}$. Therefore $H \subseteq gHg^{-1}$, so $H = gHg^{-1}$.
(ii) $\Rightarrow$ (iii): Suppose (ii) holds. Then:
(iii) $\Rightarrow$ (i) is similar (it's an exercise below). $\blacksquare$
Characterization (iii) gives a completely different way to see normality: compare the left coset and the right coset for each $g$. In $D_8$, try $H = \langle r \rangle$, then try $H = \langle s \rangle$.
Pick a group, a subgroup, and an element. The left mirror shows $gH$, the right mirror shows $Hg$.
Prove (iii) $\Rightarrow$ (i): if $gH = Hg$ for every $g \in G$, then $gHg^{-1} \subseteq H$ for every $g \in G$.
Proof. Let $g \in G$ and $h \in H$. Then $gh \in gH = Hg$, so $gh = h'g$ for some $h' \in H$. Multiplying on the right by $g^{-1}$ gives $ghg^{-1} = h' \in H$. Since $h$ was arbitrary, $gHg^{-1} \subseteq H$. $\blacksquare$
Even when $gHg^{-1}$ isn't equal to $H$, it's never far from it: it's always a subgroup, and it always has exactly the same structure as $H$.
Suppose $G$ is a group and $H$ is a subgroup of $G$. Then, $\forall g \in G$, the set $gHg^{-1}$ is also a subgroup of $G$, and the map $\phi_g : H \rightarrow gHg^{-1}$ defined by $\phi_g(h) = ghg^{-1}$ is an isomorphism.
Proof. Suppose $g \in G$, and consider the map $\tilde{\phi}_g : H \rightarrow G$, $\tilde{\phi}_g(h) = ghg^{-1}$. For any $h, h' \in H$,
$$\tilde{\phi}_g(hh') = ghh'g^{-1} = gh(g^{-1}g)h'g^{-1} = (ghg^{-1})(gh'g^{-1}) = \tilde{\phi}_g(h)\tilde{\phi}_g(h').$$
Therefore $\tilde{\phi}_g$ is a homomorphism $\Rightarrow gHg^{-1} = \tilde{\phi}_g(H) \le G$ (the image of a homomorphism is a subgroup). It follows that $\phi_g : H \rightarrow gHg^{-1}$, $\phi_g(h) = ghg^{-1}$ is a surjective homomorphism.
Finally, if $h, h' \in H$ and $\phi_g(h) = \phi_g(h')$ then $ghg^{-1} = gh'g^{-1} \Rightarrow h = h'$. Therefore $\phi_g$ is also injective.
Conclusion: $\phi_g$ is an isomorphism. $\blacksquare$
That fact is surprisingly powerful. Here's a normality proof with almost no computation at all.
$G = D_8 = \langle r, s \mid r^4 = e, s^2 = e, rs = sr^3 \rangle = \{e, r, r^2, r^3, s, sr, sr^2, sr^3\}$ and $H = \langle r \rangle = \{e, r, r^2, r^3\}$. Is $H$ normal?
Note: $\forall g \in G$, $gHg^{-1} \le G$, and $gHg^{-1} \cong H \cong C_4$. Therefore $gHg^{-1} = \langle x \rangle$ for some $x \in G$ with $|x| = 4$.
Then $x = r$ or $r^3$ (the only elements of order $4$ in $D_8$) $\Rightarrow gHg^{-1} = H$.
Conclusion: $H \unlhd G$.
The same reasoning works whenever a subgroup is one of a kind.
Any time a group $G$ contains a subgroup $H$ which is not isomorphic to any other subgroup of $G$, we have $gHg^{-1} \cong H \Rightarrow gHg^{-1} = H \Rightarrow H \unlhd G$.
In $D_8$, is $\langle s \rangle = \{e, s\}$ normal? Is $\langle r^2 \rangle = \{e, r^2\}$ normal?
$\langle s \rangle$ is not normal. From $rs = sr^3$ we get $rsr^{-1} = sr^3 \cdot r^3 = sr^6 = sr^2 \notin \langle s \rangle$.
$\langle r^2 \rangle$ is normal. Using the relation twice, $r^2 s = s r^{-2} = s r^2$, so $r^2$ commutes with $s$, and of course with $r$. So $r^2$ commutes with everything, and $g r^2 g^{-1} = r^2$ for every $g$. (Notice the shortcut from the Pro Tip doesn't apply here: $\langle r^2 \rangle \cong C_2$, just like $\langle s \rangle$. Being one of a kind is enough for normality, but not necessary.)
Here's the most common way normal subgroups show up in nature: as kernels.
If $G$ and $K$ are groups and $\phi : G \rightarrow K$ is a homomorphism, then $\ker(\phi) \unlhd G$.
Proof. We already know that $\ker(\phi) = \{g \in G : \phi(g) = e_K\}$ is a subgroup of $G$. $\forall g \in G, h \in \ker(\phi)$,
$$\phi(ghg^{-1}) = \phi(g)\phi(h)\phi(g)^{-1} = \phi(g)\phi(g)^{-1} = e_K$$
(since $\phi$ is a homomorphism). Therefore $ghg^{-1} \in \ker(\phi)$. $\blacksquare$
Use the determinant to find a normal subgroup of $GL_2(\mathbb{R})$.
The map $\phi : GL_2(\mathbb{R}) \rightarrow \mathbb{R} \setminus \{0\}$, $\phi(A) = \det(A)$, is a homomorphism with $\ker(\phi) = \{A \in GL_2(\mathbb{R}) : \det(A) = 1\} = SL_2(\mathbb{R})$.
Therefore $SL_2(\mathbb{R}) \unlhd GL_2(\mathbb{R})$.
Notice how much work that saved: checking $ABA^{-1} \in SL_2(\mathbb{R})$ by hand for every pair of matrices would be miserable.
Use the fact that kernels are normal to show that $A_n \unlhd S_n$.
The sign map $\operatorname{sgn} : S_n \rightarrow \{1, -1\}$ (sending even permutations to $1$ and odd ones to $-1$) is a homomorphism, because the parity of a product is determined by the parities of its factors. Its kernel is exactly the set of even permutations, $A_n$. Since kernels are normal, $A_n \unlhd S_n$. (This generalizes the $A_3 \unlhd S_3$ example above.)
Checking $ghg^{-1} \in H$ for every $g$ and every $h$ is a lot of work. When $G$ and $H$ come with generators, it's enough to check the generators.
If $H \le G$, $G = \langle S \rangle$, and $H = \langle T \rangle$, then:
The proof of (i) is longer, so it's tucked away below. The idea: every element is a product of generators, and a conjugate of a product is the product of the conjugates.
Proof of (i). The $\Rightarrow$ implication follows from the definition of "normal subgroup." To prove $\Leftarrow$, suppose that $\forall s \in S, t \in T$, condition $(*)$ holds.
Step 1: Show that, $\forall s \in S, h \in H$, $shs^{-1} \in H$ and $s^{-1}hs \in H$.
Since $H = \langle T \rangle$, $\forall h \in H$, $\exists t_1, \dots, t_n \in T$, $u_1, \dots, u_n \in \{\pm 1\}$ s.t. $h = t_1^{u_1}t_2^{u_2}\cdots t_n^{u_n}$. Then, $\forall s \in S$, we have:
$$shs^{-1} = st_1^{u_1}t_2^{u_2}\cdots t_n^{u_n}s^{-1} = st_1^{u_1}(s^{-1}s)t_2^{u_2}(s^{-1}s)\cdots(s^{-1}s)t_n^{u_n}s^{-1}$$
$$= (st_1^{u_1}s^{-1})(st_2^{u_2}s^{-1})\cdots(st_n^{u_n}s^{-1}) \in H$$
and
$$s^{-1}hs = s^{-1}t_1^{u_1}t_2^{u_2}\cdots t_n^{u_n}s = s^{-1}t_1^{u_1}(ss^{-1})t_2^{u_2}(ss^{-1})\cdots(ss^{-1})t_n^{u_n}s$$
$$= (s^{-1}t_1^{u_1}s)(s^{-1}t_2^{u_2}s)\cdots(s^{-1}t_n^{u_n}s) \in H$$
Step 2: Show that, $\forall g \in G, h \in H$, $ghg^{-1} \in H$.
$\forall g \in G$, $\exists s_1, \dots, s_n \in S$, $w_1, \dots, w_n \in \{\pm 1\}$ s.t. $g = s_1^{w_1}s_2^{w_2}\cdots s_n^{w_n}$. Then, $\forall h \in H$,
$$ghg^{-1} = (s_1^{w_1}s_2^{w_2}\cdots s_n^{w_n})h(s_1^{w_1}s_2^{w_2}\cdots s_n^{w_n})^{-1} = s_1^{w_1}s_2^{w_2}\cdots s_{n-1}^{w_{n-1}}s_n^{w_n}h s_n^{-w_n}s_{n-1}^{-w_{n-1}}\cdots s_2^{-w_2}s_1^{-w_1}$$
$$= s_1^{w_1}\Big(s_2^{w_2}\cdots\big(s_{n-1}^{w_{n-1}}(s_n^{w_n}h s_n^{-w_n})s_{n-1}^{-w_{n-1}}\big)\cdots s_2^{-w_2}\Big)s_1^{-w_1} \in H,$$
applying Step 1 from the innermost parentheses outward. Therefore $H \unlhd G$. $\blacksquare$
Proof of (ii). If $|G| < \infty$ the proof is essentially the same, except that we can use the fact that $\forall g \in G$, $\exists s_1, \dots, s_n \in S$ s.t. $g = s_1 \cdots s_n$. (There's no need to explicitly allow inverses of the generators.) This follows from the fact that, if $|G| < \infty$ then every element has finite order, so $\forall s \in S$, $s^{|s|} = e \Rightarrow s^{-1} = s^{|s| - 1}$ (so we can use positive powers to get inverses). $\blacksquare$
Now watch how little checking that leaves for the quaternion group.
$Q_8$ has six different square roots of $-1$: $\pm i$, $\pm j$, $\pm k$. The complex numbers only get two. Show-off.
$G = Q_8 = \{\pm 1, \pm i, \pm j, \pm k\} = \langle i, j \rangle$ and $H = \langle i \rangle = \{\pm 1, \pm i\}$. Is $H$ normal?
$Q_8$ is finite, so by part (ii) of the generator test above we only need $sts^{-1} \in H$ for $s \in \{i, j\}$ and $t = i$:
Conclusion: $H \unlhd G$.
The same argument works for every subgroup of $Q_8$, which makes $Q_8$ a genuinely strange group.
Which subgroups of $Q_8$ are normal?
Every one of them is normal. This is an example of a non-Abelian group in which every subgroup is normal.
Show that $\langle j \rangle \unlhd Q_8$ using part (ii) of the generator test. Which conjugates do you actually need to check?
With $S = \{i, j\}$ and $T = \{j\}$, only two conjugates are needed:
So $\langle j \rangle \unlhd Q_8$.
Normality asks whether conjugation moves a subgroup. We can flip the question around and ask which elements leave something alone. Suppose $G$ is a group, and let $S \subseteq G$.
The center of $G$ is $Z(G) = \{g \in G : \forall h \in G, gh = hg\}$.
The center is everything that commutes with the whole group. Shrink "the whole group" down to a set $S$, and you get:
The centralizer of $S$ in $G$ is $C_G(S) = \{g \in G : \forall s \in S, gs = sg\}$.
The centralizer commutes with each element of $S$ individually. Asking less (only that $S$ as a whole is preserved) gives:
The normalizer of $S$ in $G$ is $N_G(S) = \{g \in G : gSg^{-1} = S\}$.
A few things follow almost immediately from the definitions.
Those containments are more than just containments: every set in the chain is a subgroup of the next.
For any group $G$ and $S \subseteq G$, $Z(G) \le C_G(S) \le N_G(S) \le G$.
Proof.
$N_G(S) \le G$ (subgroup criterion):
Therefore, $N_G(S) \le G$.
$C_G(S) \le N_G(S)$: We already know that $C_G(S) \subseteq N_G(S)$, so we just need to show that $C_G(S)$ is a group.
Therefore, $C_G(S) \le N_G(S)$.
$Z(G) \le C_G(S)$: $Z(G) \subseteq C_G(S)$, and $Z(G) = C_G(G) \le G$ (by the previous part, with $S = G$), so $Z(G)$ is a subgroup sitting inside $C_G(S)$. Therefore $Z(G) \le C_G(S)$. $\blacksquare$
The quaternion group shows off all three sets nicely, because they can all be different. The explorer below sorts every element of $Q_8$ into the innermost ring it belongs to.
Toggle elements into $S$ (or pick a preset). Click any element to see why it landed where it did.
Let $Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}$. Find $Z(Q_8)$, $C_{Q_8}(\langle i \rangle)$, $N_{Q_8}(\langle i \rangle)$, $C_{Q_8}(\{i, j\})$, and $N_{Q_8}(\{i, j\})$.
Notice the second and third bullets: $C_{Q_8}(\langle i \rangle) = \langle i \rangle$ is much smaller than $N_{Q_8}(\langle i \rangle) = Q_8$. The centralizer asks every element of $S$ to stay put; the normalizer only asks $S$ to stay put as a set, so its elements can be shuffled around.
Find $Z(S_3)$ and $Z(D_8)$.
$Z(S_3) = \{e\}$. Every transposition fails to commute with some other transposition, and $(123)(12) = (13) \neq (23) = (12)(123)$ (composing right to left), so neither 3-cycle is central either.
$Z(D_8) = \{e, r^2\}$. We saw in the $D_8$ exercise above that $r^2$ commutes with everything. On the other hand, $rs = sr^3 \neq sr$, so $r, r^3, s$ are not central, and similar computations rule out $sr, sr^2, sr^3$.
One more connection back to the first half of the page: the center is always a normal subgroup.
Prove that $Z(G) \unlhd G$ for every group $G$.
Proof. By the theorem above, $Z(G) \le G$. Let $g \in G$ and $z \in Z(G)$. Since $z$ commutes with $g$, $gzg^{-1} = zgg^{-1} = z \in Z(G)$. So $gZ(G)g^{-1} \subseteq Z(G)$ for every $g \in G$, and $Z(G) \unlhd G$. $\blacksquare$
Find $C_{Q_8}(\langle j \rangle)$ and $N_{Q_8}(\{i\})$.
$C_{Q_8}(\langle j \rangle) = \langle j \rangle = \{\pm 1, \pm j\}$, by the same argument as for $\langle i \rangle$.
$N_{Q_8}(\{i\}) = \{\pm 1, \pm i\}$. For a one-element set, $g\{i\}g^{-1} = \{i\}$ means exactly $gig^{-1} = i$, so the normalizer equals the centralizer $C_{Q_8}(\{i\}) = \langle i \rangle$. For example, $jij^{-1} = -i \neq i$.
A classic, and a great showcase for characterization (iii). The coset mirror already hinted at it: in $D_8$, the subgroups $\langle r \rangle$ and $\{e, r^2, s, sr^2\}$ (both half the size of $D_8$) always passed.
Prove that every subgroup of index $2$ is normal: if $H \le G$ and $H$ has exactly two left cosets in $G$, then $H \unlhd G$.
Proof. We show $gH = Hg$ for every $g \in G$, then use characterization (iii).
If $g \in H$, then $gH = H = Hg$, since $H$ is closed under multiplication.
If $g \notin H$, then $gH \neq H$ (because $g = ge \in gH$ but $g \notin H$). The left cosets partition $G$, and there are only two of them, $H$ and $gH$, so $gH = G \setminus H$. The same argument works for right cosets: $H$ has exactly as many right cosets as left cosets (the map $xH \mapsto Hx^{-1}$ matches them up), so the two right cosets are $H$ and $Hg$, and $Hg = G \setminus H$.
Either way $gH = Hg$. By the theorem, $H \unlhd G$. $\blacksquare$
This gives another proof that $A_n \unlhd S_n$, since $A_n$ has index $2$ in $S_n$.