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The subgroups whose left and right cosets agree: conjugation as relabelling, the families that are normal without computation, and why the condition is exactly what multiplying cosets requires.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to test a subgroup for normality by conjugating, state the three equivalent forms of the condition and say why the elementwise reading is wrong, recognise the four families that are normal for free, count the normal subgroups of an abelian group, read normality as being a union of conjugacy classes, and say what the condition will be used for.
Cosets came in two kinds and an example was given where they differ: inside the permutations of three letters, the subgroup holding one transposition has $gH \ne Hg$. For most of the subgroups met so far — the even permutations, the rotations, anything inside an abelian group — the two agree. This lesson gives the agreement a name and finds out what it is good for.
The conjugate of $x$ by $g$ is $gxg^{-1}$. A subgroup $N \le G$ is normal, written $N \trianglelefteq G$, when $gN = Ng$ for every $g$, equivalently $gNg^{-1} = N$ for every $g$. The centre $Z(G)$ is the set of elements commuting with everything. A group is simple when its only normal subgroups are the trivial one and itself.
$N \le G$ is normal when any of these equivalent conditions holds:
Condition 1 does not say $gn = ng$ for each element. It says the two sets coincide, so conjugating $n$ gives some element of $N$, not necessarily $n$ itself. Reading it elementwise is the commonest error here.
Condition 3 looks weaker than 2 — it gives $gNg^{-1} \subseteq N$ rather than equality — but applying it with $g^{-1}$ in place of $g$ supplies the reverse inclusion, so the three are the same.
Normal for free. Four families need no computation:
Why it matters. Normality is exactly the condition that lets cosets be multiplied: $(aN)(bN) = abN$ is well defined precisely when $N$ is normal. That is the whole reason the definition exists, and unit 4 opens with it.
A quotient forgets on purpose. Everything in one coset of $N$ becomes a single element of $G/N$, so a question about $G$ that does not depend on $N$ can be asked in a smaller group — and the price is that questions which do depend on $N$ can no longer be asked at all.
Another way: picture
Conjugation is relabelling. In the symmetric groups it is literally that: $\sigma(a\,b\,c)\sigma^{-1} = (\sigma a\,\sigma b\,\sigma c)$, the same cycle with every letter renamed. So a subgroup is normal when it is unchanged by every relabelling the group can perform. The subgroup holding only $(1\,2)$ is not, because relabelling turns $(1\,2)$ into $(2\,3)$; the subgroup of all three transpositions together with the identity would be stable — except that it is not a subgroup, since a product of two transpositions is a three-cycle.
Another way: steps
To decide whether $N$ is normal: 1. Is the index two? Then yes, and stop. 2. Is the group abelian, or is $N$ inside the centre? Then yes, and stop. 3. Is $N$ the kernel of a map you already have? Then yes. 4. Otherwise take a general $n \in N$ and a general $g \in G$ and compute $gng^{-1}$: if it always lands in $N$, normal; one escape is enough to say no.
Conjugation $x \mapsto gxg^{-1}$ is an isomorphism of $G$ with itself: it respects the operation, since $g(xy)g^{-1} = (gxg^{-1})(gyg^{-1})$, and it is undone by conjugating by $g^{-1}$. So it preserves everything structural — orders of elements, sizes of subgroups, whether a subgroup is cyclic.
The conjugacy class of $x$ is the set of all $gxg^{-1}$. Classes partition the group, the class of a central element is the single element itself, and in $S_n$ two permutations are conjugate exactly when they have the same cycle type. So in $S_4$ the classes are the five cycle types, of sizes $1, 6, 3, 8, 6$.
Normality now has a crisp reading: $N$ is normal exactly when it is a union of conjugacy classes. That gives a fast way to find the normal subgroups of a small group. In $S_4$, a normal subgroup must contain the identity and be a union of classes whose total size divides $24$: the possibilities are $1$, $1 + 3 = 4$, $1 + 3 + 8 = 12$ and $24$, and each of those does occur. Nothing of size $6$ can be assembled, which is the counting proof of last lesson's counterexample.
This reading also explains what a simple group is: one whose conjugacy classes cannot be assembled into anything but the trivial subgroup and the whole group. $A_n$ for $n \ge 5$ is simple, and that fact is why the general polynomial equation of degree five has no formula in radicals — a result belonging to Abstract algebra II but standing on exactly this definition.
Reading $gN = Ng$ elementwise. It says the two sets are equal, not that $gn = ng$ for each $n$. A normal subgroup of a non-abelian group is not made of central elements.
Thinking normal means abelian, or means the group is abelian. Neither. The even permutations of four letters form a normal subgroup that is not abelian, inside a group that is not abelian.
Assuming normality is transitive. It is not: $N \trianglelefteq H$ and $H \trianglelefteq G$ do not give $N \trianglelefteq G$. The standard counterexample lives inside the symmetries of a square.
Checking only one $g$. The condition is for every $g$ in the group. One conjugate landing inside proves nothing; one landing outside proves the subgroup is not normal.
Expecting a subgroup of a normal subgroup's order to be normal. Order has nothing to do with it, except in the single case of index two.
In the permutations of three letters take $H = \{e, (1\,2)\}$, of order $2$ and index $3$.
Index three, so nothing is free.
Conjugate: $(1\,3)(1\,2)(1\,3)^{-1} = (2\,3)$, which is not in $H$.
One escape is enough.
So $H$ is not normal. Seen as relabelling: conjugating by $(1\,3)$ renames $1$ as $3$, turning the swap of $1$ and $2$ into the swap of $3$ and $2$.
Conjugation renames.
The four rotations form a subgroup of order $4$ inside a group of order $8$, so the index is $2$.
Index two.
So it is normal, with no conjugation computed. Checking one case anyway: a reflection conjugates a quarter turn to the three-quarter turn, still a rotation.
The free reason, confirmed.
The subgroup generated by a single reflection has order $2$ and index $4$, and it is not normal: conjugating a reflection by a quarter turn gives a different reflection.
Same group, two subgroups, two answers.
Take $z$ in the centre and any $g$. By definition $zg = gz$.
Use the defining property.
So $gzg^{-1} = zgg^{-1} = z$, which is in the centre.
Every conjugate lands back inside.
So the centre is normal — and more than that, every element is fixed by conjugation rather than merely staying inside. Any subgroup of the centre is normal for the same reason.
Select every subgroup that is normal in the group it sits in.
This task has no paper form; do it on a device.
Build the proof that a subgroup of index two is normal.
This task has no paper form; do it on a device.
Say how many normal subgroups each of these groups has, counting the trivial subgroup and the whole group.
| Normal subgroups | |
|---|---|
| The integers modulo 12 under addition | |
| The permutations of three letters | |
| A group of prime order | |
| The trivial group |
Each of these subgroups is normal for a reason that needs no computation. Match it to the reason.
| it commutes with everything | it has index two | conjugation in the ambient group does nothing | it is the kernel of a homomorphism | it is cyclic | |
|---|---|---|---|---|---|
| The even permutations, inside all the permutations | |||||
| The centre of a group | |||||
| Any subgroup of an abelian group | |||||
| The matrices of determinant one, inside the invertible matrices |
Is the rotations by a multiple of a third turn inside the symmetries of a triangle normal?
How many normal subgroups does the integers modulo $12$ under addition have?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Is every subgroup of an abelian group normal?
You can decide whether a subgroup is normal, and you know which subgroups are normal without any checking. Say in your own words why $gN = Ng$ does not mean the elements of $N$ commute with $g$. Next: homomorphisms, the maps that respect the operation — and the source of every normal subgroup there is.
9. Your turn: is the centre of a group always normal?, step 3