Products and the Isomorphism Theorems

Section 3.4 Products and the Isomorphism Theorems

Subsection The First Isomorphism Theorem

“If it’s your job to eat a frog, it’s best to do it first thing in the morning. And If it’s your job to eat two frogs, it’s best to eat the biggest one first.”
―Mark Twain

We come to the so-called Isomorphism Theorems.

Theorem 3.42. UMP for Quotient Groups.

Let \(G\) be a group and \(N\) a normal subgroup. If \(\varphi: G \to H\) is a homomorphism of groups such that \(N \subseteq \operatorname{ker}(\varphi)\text{,}\) then

there exists a unique group homomorphism \(\overline{\varphi}: G/N \to H\) such that the composition of \(\overline{\varphi}\) and the quotient map \({\pi}: G\twoheadrightarrow G/N\) is \(\varphi\text{.}\)
If \(\varphi\) is onto, then \(\overline{\varphi}\) is onto.
Moreover, \(\operatorname{ker}(\overline{\varphi}) = \operatorname{ker}(\varphi)/N = \{ gN \mid \varphi(g) = e_H\}.\)

Proof.

If such a \(\overline{\varphi}\) exists, it is necessarily unique since \(G \twoheadrightarrow G/N\) is onto. In fact, if \(\varphi=\pi\circ \overline{\varphi}\) then \(\overline{\varphi}\) has to be given by the formula

\begin{equation*} \overline{\varphi}(gN) = \varphi(g). \end{equation*}

We now need to show that this formula determines a well-defined homomorphism: if \(xN = yN\text{,}\) then \(y^{-1}x \in N \subseteq \operatorname{ker}(\varphi)\) and so \(\varphi(y)^{-1} \varphi(x) = e\text{,}\) whence \(\varphi(y) = \varphi(x)\text{.}\) For any \(x,y\) we have

\begin{equation*} \overline{\varphi}(xNyN) = \overline{\varphi}(xyN) = \varphi(xy) = \varphi(x)\varphi(y) =\overline{\varphi}(xN) \overline{\varphi}(yN). \end{equation*}
The formula for \(\overline{\varphi}\) given above ensures that \(\operatorname{im}\varphi=\operatorname{im}\overline{\varphi}\) hence \(\varphi\) is surjective if and only if \(\overline{\varphi}\) is surjective.
We have \(\overline{\varphi}(xN) = e_H\) iff \(\varphi(x) = e_H\) iff \(x \in \operatorname{ker}(\varphi)\) iff \(xN \in \operatorname{ker}(\varphi)/N\text{.}\) If \(xN \in \operatorname{ker}(\varphi)/N\) for some \(x \in G\text{,}\) then \(xN = yN\) for some \(y \in \operatorname{ker}(\varphi)\) and hence \(x = yz\) for some \(z \in N\text{.}\) Since \(N \subseteq \operatorname{ker}(\varphi)\text{,}\) we have \(x \in \operatorname{ker}(\varphi)\text{.}\)

Example 3.43. Abelianization.

Let \(G\) be any group, and recall the Commutator Subgroup. In Example 3.19 we saw \(G'\nsg G\text{.}\) Now let \(\varphi: G \to A\) be any group homomorphism from \(G\) to an abelian group \(A\text{.}\) Since \(\varphi([x,y]) = [\varphi(x),\varphi(y)] = e\) for all \(x, y \in G\) (since \(A\) is abelian), we have that \(\operatorname{ker}(\varphi)\) must contain \(G'\text{.}\) By Theorem 3.42, we get that \(\varphi\) factors as

\begin{equation*} \varphi:G \xrightarrow{{\pi}} G/G' \xrightarrow{\overline{\varphi}} A \end{equation*}

for a unique group homomorphism \(\overline{\varphi}\text{.}\)

The group \(G/G'\) is called the abelianization of \(G\) and the motto is: a homomorphism from a group to an abelian group factors uniquely through the abelianization.

Theorem 3.44. First Isomorphism Theorem.

If \(\varphi: G \to H\) is a homomorphism of groups, then \(\operatorname{ker}(g)\mathrel{\unlhd}G\) and the map \(\overline{f}\) defined by Theorem 3.42 induces an isomorphism

\begin{equation*} \overline{\varphi}: G/\operatorname{ker}(\varphi) \xrightarrow{\cong} \operatorname{Im}(\varphi). \end{equation*}

Proof.

By Theorem 3.42, there exists a homomorphism \(\overline{f}\) such that \(\overline{f} \circ {\pi}= f\text{,}\) and its kernel consists of just the one element \(\operatorname{Ker}(f)/\operatorname{Ker}(f)\) of \(G/\operatorname{Ker}(f)\text{.}\) So \(\overline{f}\) is one-to-one, and the image of \(\overline{f}\) is the same as the image of \(f\text{.}\)

Corollary 3.45. Double Divide.

If \(\varphi\) is a homomorphism from a finite group \(G\) to a group \(H\text{,}\) then \(|\varphi(G)|\) divides \(|G|\) and \(|H|\text{.}\)

Exercise 3.46. \(G/Z(G)\cong\Inn(G)\).

For any group \(G\) we have \(G/Z(G)\cong\Inn(G)\text{.}\)

Subsection Products

“Outside of a dog, a book is a man’s best friend. Inside of a dog it’s too dark to read.”
―Groucho Marx

Exercise 3.47. Trivially Intersecting Normal Subgroups.

Let \(H\) and \(K\) be normal subgroups of a group \(G\) such that \(H\cap K=\{1\}\text{.}\) Prove that \(xy=yx\) for all \(x\in H, y\in K\text{.}\)

Definition 3.48. Group Product (\(HK\)).

Let \(H\) and \(K\) be subgroups of a group and define the set

\begin{equation*} HK = \{hk \mid h \in H, k\in K\}. \end{equation*}

Warning 3.49. Product vs. Direct Product.

The product and direct product

Whatever that is

are not always the same. We will explore this later in Chapter 7

Theorem 3.50. Product Order.

For two finite subgroups \(H,K\leq G\text{,}\) \(|HK|=\frac{|H|\cdot|K|}{|H\cap K|}\text{.}\)

Theorem 3.51. Products and Normality.

Let \(G\) be a group, \(H \leq G\) and \(N \nsg G\text{.}\) Then

\(HN \leq G\text{,}\)
\(N \cap H \nsg H\text{,}\) and
\(N \nsg HN\text{.}\)

Corollary 3.52. When \(HK=KH\).

If either one of \(H\) or \(K\) is a normal subgroup of \(G\text{,}\) then \(HK=KH=\langle H\cup K\rangle\text{.}\)

Warning 3.53.

The identity \(HK=KH\) does not mean that every pair of elements from \(H\) and \(K\) must commute.

Example 3.54. \(HK=KH\) but not Abelian.

In \(D_{2n}\text{,}\) let \(H=\langle s\rangle\) and \(K=\langle r\rangle\text{.}\) Then \(HK=KH=D_{2n}\) but \(r\) and \(s\) do not commute. The fact that \(HK=KH\) can also be justified by observing that \(K\nsg D_{2n}\text{.}\)

The second isomorphism theorem, also known as the diamond isomorphism theorem or the modular law, provides information about the structure of subgroups and their intersections.

Theorem 3.55. Second Isomorphism Theorem.

Let \(G\) be a group, \(H \leq G\) and \(N \unlhd G\text{.}\) Then there is an isomorphism

\begin{equation*} \frac{H}{N \cap H} \xrightarrow{\cong} \frac{HN}{N} \end{equation*}

given by

\begin{equation*} h \cdot (N \cap H) \mapsto hN. \end{equation*}

Proof.

The first two assertions are left as exercises and since \(N \unlhd G\) we have \(N \unlhd HN\text{.}\) Define a homomorphism \(\varphi: H \to \frac{HN}{N}\) by \(\varphi(h) = hN\text{.}\) This is a homomorphism since it is the composition [provisional cross-reference: cite]

\begin{equation*} \varphi:H \hookrightarrow HN \xrightarrow{{\pi}} \frac{HN}{N} \end{equation*}

of homomorphisms. \(\varphi\) is onto since for all \(h,n\) we have \(hnN = hN = \varphi(h)\text{.}\) The kernel of \(\varphi\) is \(\operatorname{ker}(\varphi)=\{h \mid hN = N\} = H \cap N\text{.}\) The result thus follows from the First Isomorphism Theorem.

The third isomorphism theorem, also known as the factor or quotient theorem, gives a relationship between normal subgroups of a group and their quotient groups.

Theorem 3.56. Third Isomorphism Theorem.

Suppose \(G\) is a group, \(M \leq N \leq G\text{,}\) \(M \unlhd G\) and \(N \unlhd G\text{.}\) Then \(M \unlhd N\text{,}\) \(N/M \unlhd G/M\) and there is an isomorphism

\begin{equation*} (G/M)/(N/M) \xrightarrow{\cong} G/N \end{equation*}

given by sending the coset of \((G/M)/(N/M)\) represented by \(gM\) to \(gN\text{.}\)

Proof.

The first two assertions are immediate from the definitions.

The kernel of the canonical map \({\pi}: G \twoheadrightarrow G/N\) contains \(M\) and so by Theorem 3.42 we get an induced homomorhism

\begin{equation*} \phi: G/M \to G/N \end{equation*}

with \(\phi(gM) = {\pi}(g) = gN\text{.}\) Moreover, we know

\begin{equation*} \operatorname{Ker}(\phi) = \operatorname{Ker}({\pi})/M = N/M. \end{equation*}

Finally apply the First Isomorphism Theorem to \(\phi\text{.}\)

Theorem 3.57. Lattice Isomorphism Theorem.

Let \(G\) be a group and \(N\) a normal subgroup with canonical homomorphism \(\pi:G\to G/N\text{.}\) There is an containment-preserving bijection

\begin{equation*} \Psi:\{\text{subgroups of}\; G \;\text{that contain}\; N\} \xrightarrow{\text{bijective}}\{\text{subgroups of}\; G/N\} \end{equation*}

given by \(\Psi(H)= H/N\) for \(N \leq H \leq G\text{.}\) The inverse is defined for \(\mathcal{H}\leq G/N\) by

\begin{equation*} \Phi(\mathcal{H})=\pi^{-1}(\mathcal{H}) = \{x \in G \mid \pi(x) \in \mathcal{H}\} \end{equation*}

where \(\pi: G \twoheadrightarrow G/N\) is the quotient map. We denote \(\Psi(H)=H/N=\overline{H}\text{.}\)

Then this bijection enjoys the following properties:

(normal) subgroups correspond to normal subgroups i.e.,
- \(H\leq G\) iff \(\overline{H}\leq \overline{G}\) and \(\mathcal{H}\leq \overline{G}\) iff \(\Phi(\mathcal{H})\leq G\)
- \(H\mathrel{\unlhd}G\) iff \(\overline{H}\mathrel{\unlhd}\overline{G}\) and \(\mathcal{H}\mathrel{\unlhd}\overline{G}\) iff \(\Phi(\mathcal{H})\mathrel{\unlhd}G\)
indices are preserved; i.e., \([G:H] = [\overline{G} : \overline{H}]\) and \([G: \Phi(\mathcal{H})] = [\overline{G} : \mathcal{H}]\text{.}\)
the supremums and infimums are preserved (this makes \(\Phi\) and \(\Psi\) lattice isomorphisms)
- \(\overline{H} \cap \overline{K} = \overline{H \cap H}\) and \(\langle \overline{H}\cup \overline{K} \rangle = \overline{\langle H\cup K \rangle}\)
- \(\Phi(\mathcal{H}) \cap \Phi(\mathcal K) = \Phi(\mathcal{H}\cap \mathcal K)\) and \(\langle \Phi(\mathcal{H})\cup\Phi(\mathcal K) \rangle = \Phi\left(\langle \mathcal{H}\cup\mathcal K\rangle\right)\)

Proof.

We have previously shown Proposition 3.39 that the quotient map \(\pi:G\to G/N\) is a surjective group homomorphism. We show:

\(\Psi\) is well defined (correct codomain) since for \(H\leq G\) we have \(\pi(H)\leq G/N\) (since images of subgroups through group homomorphisms are subgroups by Theorem 1.100).
\(\Phi\) is well defined (correct codomain) since for \(\mathcal{H}\leq G\) we have \(\pi^{-1}(\mathcal{H})\leq G\) (by Theorem 1.100 again) and for any \(\mathcal{H}\leq G\) we have \(\{e_GN\}\subseteq \mathcal{H}\text{,}\) hence

\begin{equation*} N=\operatorname{Ker}(\pi)=\pi^{-1}(\{e_GN\})\subseteq \pi^{-1}(\mathcal{H})=\Phi(\mathcal{H}). \end{equation*}
\(\Phi\) and \(\Psi\) are mutual inverses: \((\Psi\circ\Phi)(\mathcal{H})=\pi(\pi^{-1}(\mathcal{H}))=\mathcal{H}\) since \(\pi\) is surjective and \((\Phi\circ\Psi)(H)=\pi^{-1}(\pi(H))=\pi^{-1}(H/N)=H\text{,}\) with the last equality justified by

\begin{equation*} \begin{aligned} x\in \pi^{-1}(H/N) &\iff \pi(x)\in H/N \iff xN=hN \text{ for some } h\in H \\ & \iff x\in hN \text{ for some } h\in H \iff x\in H \text{ (using that } N\subseteq H).\\ \end{aligned} \end{equation*}

Thus, the two functions defined in the statement are well-defined and are mutually inverse.

Since \(\pi\) and \(\pi^{-1}\) preserve containments, each of \(\Psi\text{,}\) \(\Psi^{-1}\) preserves the order relation of containment.

I will only prove some parts of statements (1), (2), (3) in the theorem.

If \(N \leq H \leq G\) and \(H \unlhd G\text{,}\) then \(H/N \unlhd G/N\) holds by part of Theorem 3.56 or by the exercise below, since \(\pi\) is surjective. The fact that the inverse function also sends normal subgroups to normal subgroups is a consequence of the statement that inverse images of normal subgroups are normal subgroups.
²
See: Proposition 3.22.
In the interest of time, I’ll only prove the assertion about indices in the special case when \(H\) is normal. In that case this fact is also an immediate consequence of the Third Isomorphism Theorem since for \(N \leq H \leq G\) with \(H \unlhd G\) we have

\begin{equation*} [G:H] = |G/H| = \left|(G/N)/(H/N)\right| = [G/N: H/N]=[\overline{G}:\overline{H}]. \end{equation*}

The general case is a consequence of an exercise from HW 5. [provisional cross-reference: cite]
The proof of (3) is omitted.

Summary

The First Isomorphism Theorem tells us \(G/\operatorname{ker}(\varphi)\cong\operatorname{Im}(\varphi)\) for any homomorphism \(\varphi\text{.}\)
For \(H,K\leq G\text{,}\) we define the Group Product (\(HK\)) to be the set \(HK = \{hk \mid h \in H, k\in K\}\text{.}\)
\(HK\leq G\) if either \(H\) or \(K\) are normal. If \(N\nsg G\text{,}\) then \(N \cap H \nsg H\) and \(N \nsg HN\text{.}\)
³
See: Theorem 3.51.
We also have the Second Isomorphism Theorem, Third Isomorphism Theorem, and Lattice Isomorphism Theorem.

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