Normal subgroups, quotients, isomorphism theorems

In one sentence

Normal subgroups are exactly the kernels, quotients exist exactly for them, and the property that catches people is that normality is a relation between two groups rather than something a subgroup possesses.

Why the exam asks it

Because "normal" gets treated as an adjective attached to a subgroup, and it is not — it is always normal in something. That misreading produces the transitivity error, which is set often and has a small concrete counterexample.

The idea

$N \le G$ is normal, written $N \trianglelefteq G$, if $gNg^{-1} = N$ for every $g \in G$; equivalently every left coset is a right coset.

Normality is exactly what makes the coset multiplication $(aN)(bN) = abN$ well defined, so the quotient group $G/N$ exists precisely for normal $N$.

Sufficient conditions worth recognising:

• Index $2$ $\Rightarrow$ normal.
• Any subgroup of an abelian group.
• The centre $Z(G)$, and more generally any subgroup of the centre.
• Kernels: $\ker\varphi \trianglelefteq G$ for every homomorphism, and conversely every normal subgroup is the kernel of the quotient map. Normal subgroups and kernels are the same objects.

The isomorphism theorems

First. $G/\ker\varphi \cong \operatorname{im}\varphi$. This is the one that does the work: to identify a quotient, find a homomorphism with that kernel and look at its image.

Second. For $H \le G$ and $N \trianglelefteq G$: $HN/N \cong H/(H\cap N)$.

Third. For $N \le M$ both normal in $G$: $(G/N)/(M/N) \cong G/M$.

Correspondence. Subgroups of $G/N$ correspond exactly to subgroups of $G$ containing $N$, and normality is preserved both ways. This is how questions about a quotient's subgroup structure are answered without constructing anything.

Normality is not transitive

In $D_4 = \langle r,s : r^4 = s^2 = e,\; srs = r^{-1}\rangle$ of order $8$:

• $H = \{e, r^2, s, r^2s\}$ has index $2$, so $H \trianglelefteq D_4$.
• $N = \{e, s\} \le H$, and $H$ is abelian, so $N \trianglelefteq H$.
• But $rsr^{-1} = r^2 s \notin N$, so $N$ is not normal in $D_4$.

$N \trianglelefteq H \trianglelefteq G$ does not give $N \trianglelefteq G$. The reason is visible in the definition: normality of $N$ in $H$ constrains conjugation by elements of $H$ only, and $G$ has more elements.

The property that is transitive is being characteristic — preserved by every automorphism. Characteristic in normal gives normal, which is the correct repair.

Simple groups and solvability

$G$ is simple if its only normal subgroups are $\{e\}$ and $G$. The abelian simple groups are exactly $C_p$. The smallest non-abelian simple group is $A_5$, of order $60$, and $A_n$ is simple for every $n \ge 5$ — which is what makes the general quintic unsolvable.

The commutator subgroup $[G,G]$ is the smallest normal subgroup with abelian quotient: $G/N$ is abelian $\iff [G,G] \le N$. So $G/[G,G]$ is the largest abelian quotient.

Where intuition breaks

"$N \trianglelefteq H$ and $H \trianglelefteq G$, so $N \trianglelefteq G$." $D_4$ as above. The topic's central error, and it comes from reading "normal" as a property rather than a relation.

"$G/N$ abelian and $N$ abelian, so $G$ abelian." $S_3$ with $N = A_3$: both $A_3$ and $S_3/A_3 \cong C_2$ are abelian, and $S_3$ is not. An abelian quotient of an abelian subgroup says the group is solvable, not abelian.

"A quotient of a non-abelian group is non-abelian." $S_3/A_3 \cong C_2$. Quotients lose information, which is the point of them.

"Every subgroup is a kernel." Only the normal ones. $\langle(1\,2)\rangle$ in $S_3$ is not the kernel of anything.

"$G/N \cong H$ means $G \cong N \times H$." Far from it. $C_4/C_2 \cong C_2$ and $C_4 \not\cong C_2\times C_2$. A quotient being a group does not split the extension.

"Normal subgroups commute elementwise with everything." $gN = Ng$ as sets, not $gn = ng$ for each $n$. Elementwise commuting is the centre, which is a much stronger condition.

"$A_4$ is simple because $A_5$ is." $A_4$ has the Klein four-group as a normal subgroup. Simplicity of $A_n$ starts at $n = 5$.

The exam's angle

1. To identify a quotient, build a homomorphism. The first isomorphism theorem turns "what is $G/N$" into "what does a map with kernel $N$ hit", which is usually visible.
2. Check normality by conjugating a generator. One element that escapes settles it, as $rsr^{-1}$ does above.
3. Index 2, centre, kernel — all normal for free. Recognising one of these is often the whole step.
4. For transitivity claims, look for the $D_4$ shape: a reflection inside a Klein subgroup.
5. Use the correspondence theorem for questions about subgroups of a quotient rather than constructing cosets.

The night before

• $N \trianglelefteq G \iff gNg^{-1} = N \iff$ every left coset is a right coset $\iff$ $N$ is a kernel.
• Index $2$, the centre, and any subgroup of an abelian group are normal.
• First isomorphism theorem: $G/\ker\varphi \cong \operatorname{im}\varphi$. Second: $HN/N \cong H/(H\cap N)$. Third: $(G/N)/(M/N)\cong G/M$.
• Normality is not transitive: $\{e,s\} \trianglelefteq \{e,r^2,s,r^2s\} \trianglelefteq D_4$, and $rsr^{-1} = r^2s$. Characteristic-in-normal is the repair.
• $N$ and $G/N$ abelian does not give $G$ abelian — $S_3$ and $A_3$. It gives solvable.
• $G/N$ abelian $\iff [G,G] \le N$.
• Simple: only $\{e\}$ and $G$ normal. Abelian simple $=$ $C_p$. $A_n$ simple for $n\ge5$; $|A_5| = 60$ is the smallest non-abelian one. $A_4$ is not simple.