§11 Short Exact Sequences and Semidirect Products

§11.1 Extensions—Short Exact Sequences

Definition 11.1 A short exact sequence of groups is a sequence consisting of two homomorphisms

satisfying the following conditions:

(1) $G\to H$ is injective,

(2) $H\to K$ is surjective, and

(3) the kernel of $H\to K$ is equal to—not merely isomorphic to—the image of $G\to H$.

A short exact sequence is usually written as

Definition 11.2 We also say that $H$ is an extension of $K$ by $G$.

Question: What do the $1$'s at the two ends mean?

The symbol $1$ denotes the trivial group with one element.

The sequence above is “exact” because the image of each homomorphism is the kernel of the next homomorphism.

For example, the portion

means that the image of

is the kernel of

which means that

is injective.

For reasons that will become clearer later, short exact sequences are important because of the following idea:

we think of the group $H$ as being built out of the groups $G$ and $K$.

Example 11.1 We have the following short exact sequences:

(1)

where the first homomorphism sends

and the second homomorphism sends

and

(2)

We take

and

Thus we may think of both

and the Klein four group as being built from two copies of

But we see that different groups can be constructed from $\mathbb{Z}/2\mathbb{Z}$ in different ways.

We also have the following short exact sequences:

(3)

where

and this is the short exact sequence associated with the inclusion

and

(4)

Thus we see that there are at least two different ways to construct a group of order $6$ from

and

Remark The examples above show that:

(1) an extension need not be a direct product;

(2) an extension

need not admit a map

such that

(3) an extension of Abelian groups can be non-Abelian.

§11.2 Split Short Exact Sequences

Definition 11.3 If there exists a homomorphism

such that

we say that the short exact sequence

splits.

For readability, from now on we will write a short exact sequence

as

where $L$ stands for “left” and $R$ stands for “right”.

Since

is injective, from now on we will identify $L$ with its image in $H$ in order to simplify notation.

Example 11.2 Among the short exact sequences above:

(1)

(2)

(3)

(4)

Only (1) does not split.

Notice that, a priori, there is no reason to regard $R$ as a subgroup of $H$.

For example, in

the second copy of $\mathbb{Z}/2\mathbb{Z}$ cannot naturally be “embedded” back into $\mathbb{Z}/4\mathbb{Z}$.

Proposition 11.1

does not split.

Proof: $\mathbb{Z}/2\mathbb{Z}$ contains only elements of order $1$ and $2$.

Therefore, no homomorphism

can have an image containing an element of order $\geqslant3$.

Indeed, if

then we must also have

But in $\mathbb{Z}/4\mathbb{Z}$, both $[1]$ and $[3]$ have order $4$:

Therefore the image of any homomorphism

must lie inside

But this is the kernel of the map

Therefore no $j$ can split the identity map on

Here is a more dramatic example.

Example 11.3 The short exact sequence

does not split for any

Any homomorphism from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}$ would have to send an element of order $n$ to an element of finite order.

But $\mathbb{Z}$ has no nonzero elements of finite order.

Therefore there is no injection

Since this is our first attempt to understand short exact sequences, let us analyze the case in which $R$ can be regarded as a subgroup of $H$.

If both $L$ and $R$ live inside $H$, then perhaps it is easier to accept the idea that $H$ is “constructed” from $L$ and $R$.

This leads us to the following definition.

Definition 11.4 If there exists a group homomorphism

such that the composite

equals

then we say that the short exact sequence splits.

A choice of

is called a splitting.

Thus, if the short exact sequence is given by homomorphisms

then the definition means

In particular, $j$ is injective.

§11.3 Semidirect Products

In the example above, it is clear that

cannot be regarded as a subgroup of $\mathbb{Z}$.

So we arrive at a new idea.

We would like to be able to recognize semidirect products naturally, and we would also like to produce examples.

Let us analyze the situation.

As above, in a split short exact sequence, identify $R$ with $j(R)$.

Then every element of $R$ defines an action on $H$ by conjugation:

Since $L$ is normal,

so this defines an action on $L$:

Moreover, this is a group isomorphism from $L$ to itself.

We can prove that this defines a group homomorphism

given by

Equivalently, it is enough to prove

Indeed, for $h\in R$,

Using associativity of the group operation,

This new map

is called the conjugation action of $R$ on $L$.

Thus every splitting gives rise to a group homomorphism

Here $\mathrm{Aut}(L)$ means the automorphism group of $L$ as a group, not merely as a set.

Question: Fix two groups $R$ and $L$.

A natural question is:

does every group homomorphism

produce a split exact sequence?

Another observation is that, given a splitting, both $R$ and $L$ become subgroups of $H$.

Moreover, their intersection contains only

Indeed, if a nonidentity element

existed, then the map

could not be injective.

Finally, since the orbits of the action of $L$ cover all of $H$, we obtain

That is,

Definition 11.5 Let $L$ and $R$ be subgroups of $H$. We define

To prove that

when the short exact sequence

splits, we need the following lemma.

Lemma 11.2 Let

be a short exact sequence.

Let

be the quotient homomorphism

Then there exists an isomorphism

such that

That is, there exists $z$ such that the diagram

commutes.

Proof: We are given a short exact sequence

where

is surjective and

We want to prove that there exists an isomorphism

such that

where

is the quotient map.

Since $\psi$ is surjective and

the First Isomorphism Theorem gives

Define

by

This is well defined because if

then

so

The map $z$ is an isomorphism because:

• $z$ is a homomorphism since $\psi$ is a homomorphism;

• $z$ is surjective since $\psi$ is surjective;

• $z$ is injective because the kernel of $z$ is precisely $L$, meaning that the only element mapped to the identity of $R$ is $L$.

For every $h\in H$,

so

Therefore $z$ is the required isomorphism making the diagram commute.

By the First Isomorphism Theorem, there exists an isomorphism

such that

which proves the lemma.

Once we have this lemma, we can prove the following corollary.

Corollary 11.3 If

is a splitting of

then

Proof: By the definition of a splitting,

On the other hand, by the lemma,

Thus

Since $z$ is a group isomorphism, its inverse is a homomorphism, and therefore

Now let us interpret

The homomorphism $q$ sends $h$ to

Therefore,

sends $r$ to the coset

Since

is a bijection, for every coset

there exists a unique $r\in R$ such that

Since

this proves

In the discussion above, we used $j$ to identify $r\in R$ with its image in $H$, so we write

Question: Fix

If

$L\subset H$ is normal, and

is $H$ the semidirect product of $L$ and $R$?

This is exactly the right question, because the answer is yes.

Theorem 11.4 Fix a normal subgroup

and suppose

The following conditions are equivalent:

(1) A homomorphism

splits the short exact sequence

(2) An isomorphism

onto a subgroup

such that

and the map

is surjective.

(3) A group homomorphism

Among these descriptions, we like the last one best because it makes no reference to the group $H$.

Once we construct a group homomorphism

we can construct a short exact sequence

Question: What is the group operation on $H$ in terms of $R$ and $L$?

Proposition 11.5 Any homomorphism

defines a group $H$ and a split short exact sequence

such that:

(1) the following defines a group structure on the set

The right-hand side is almost the usual group operation on $L\times R$, except that before multiplying $l_1$ and $l_2$, we “twist” $l_2$ into another element of $L$, namely

the value of $r_1$ under the homomorphism

Moreover,

(2) the set

is a normal subgroup isomorphic to $L$;

(3) the set

is a subgroup isomorphic to $R$.

Proof: Associativity:

The third equality holds because

is a homomorphism.

The fourth equality holds because

is a homomorphism.

Identity:

The second equality follows because $\phi$ is a homomorphism, so

Inverse:

Proposition 11.6

Proof:

The second equality follows because

is a homomorphism, so

The third equality follows because $\phi$ is a group homomorphism, so

Also,

The second equality holds because

is a group homomorphism.

The final equality holds because $\phi_{r^{-1}}$ is a group homomorphism.

Therefore, this really does define a group.

Definition 11.6 We denote this group by

When $\phi$ is understood, we write

and call $L\rtimes R$ a semidirect product of $L$ and $R$.

Remark We say “a” semidirect product because different choices of $\phi$ may produce different groups.

Although the word “and” is usually insensitive to order, $L$ and $R$ play very different roles here.

Remark Why do we use the symbol

Usually, when $N$ is a normal subgroup of $G$, we write

The symbol

is a hybrid of

for “normal subgroup” and

for “product”.

Any diagram

gives rise to a map

How?

By conjugation, since

is a group automorphism of $L$.

By essentially the same argument as in the exercise, we obtain a homomorphism

The composite

is exactly the homomorphism $\phi$.

Question: How does $R$ act on

Proposition 11.7

In other words, conjugation by $r$ in

recovers $\phi_r$.

Proof:

We now have enough ingredients to prove Theorem 11.4.

The key ideas are:

• If

is normal, then

by conjugation.

• Given a splitting

then $R$ also acts by conjugation:

• $L\rtimes R$ is a group in which the conjugation action of $R$ on $L$ agrees with $\phi$.

Now let us prove

To see why

we need a lemma.

Lemma 11.8 For every

there exist unique

and

such that

Proof: We know that the diagram

commutes.

That is,

Given a splitting

we have

Since $z$ is an isomorphism, it has an inverse

and therefore

The map $z^{-1}$ is also a group isomorphism.

Since $z^{-1}$ is a bijection, for every $h\in H$ there exists a unique $r$ such that

Hence there exists a unique $r\in R$ such that

Equivalently, there exists a unique $r\in R$ such that

Thus there exists a unique $r\in R$ such that

for some $l\in L$.

Of course, once $h$ and $j(r)$ are fixed, $l$ is uniquely determined:

Therefore, for every $h\in H$, there exist unique $l,r$ such that

Now we can prove the following.

Theorem 11.9 Let

be a split short exact sequence, and let

be the induced action.

Then

Proof: Consider the map

Then

But by definition,

Therefore,

Thus $\alpha$ is a homomorphism.

By Lemma 11.8, for every $h\in H$ there exist unique $l\in L$ and $r\in R$ such that

Therefore $\alpha$ is a bijection.

This is enough to show that a split short exact sequence and a semidirect product contain the same amount of data.

• Given

we obtain

by conjugation.

• By the theorem,

is an isomorphism.

• Hence we have a surjection

• Since $\alpha$ is an isomorphism,

by Lemma 11.8.

• Therefore we obtain a short exact sequence

with splitting

$(1_{L},r)$$\mapsto$$~r$

The situation can be summarized by saying that the following diagram commutes:

That is, every sub-square that can be drawn is commutative.

Example 11.4 Recall that

is a subgroup of index $2$.

By definition,

Every kernel is a normal subgroup, so

is normal.

Alternatively, one may use the fact that every subgroup of index $2$ is normal.

Therefore we have a short exact sequence

$\mathbb{Z}/2\mathbb{Z}$ $\cong$ $1\to SO_{n}(\mathbb{R})\to O_{n}(\mathbb{R})\to{\pm 1}\to 1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ $\subset$ $\mathbb{R}^{\times}$

This sequence admits many different splittings.

For concreteness, take

For example,

$\begin{pmatrix} 1&0\ 0&1 \end{pmatrix}$$\mapsto$$[0]$

$\begin{pmatrix} 1&0\ 0&-1 \end{pmatrix}$$\mapsto$$[1]$

or

$\begin{pmatrix} -1&0\ 0&1 \end{pmatrix}$$\mapsto$$[1]$

or

$\begin{pmatrix} 0&1\ 1&0 \end{pmatrix}$$\mapsto$$[1]$