Subgroups

Chapter 6 Subgroups

Section 6.1 Definitions and Basic Examples

Definition 6.1. Subgroup.

Assume \((G,\ast)\) is a group and \(H\) is a nonempty subset of \(G\text{.}\) We say that \(H\) is a subgroup of \(G\) and write \(H \leq G\) if \(H\) is closed under \(\ast\) and \((H, \ast)\) is a group. In this case, we write \(H \leq G\text{.}\)

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If \(H\) is a subgroup of \(G\) and \(H \neq G\text{,}\) then we say \(H\) is a proper subgroup and write \(H \lt G\text{.}\)

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Definition 6.2. Trivial Subgroup.

Every group contains at least one proper subgroup, called the trivial subgroup, that contains only the identity element.

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Example 6.3. Additive Subgroups of Numbers.

The set inclusions \(\Z \subseteq \Q \subseteq \R \subseteq \C\) translate naturally into additive subgroups

\begin{equation*} \Z \leq \Q \leq \R \leq \C\text{.} \end{equation*}

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Example 6.4. Multiplicative Subgroups of \(\C^\times\).

The set inclusions \(U_n \subseteq U \subseteq \C^\times\) translate naturally into multiplicative subgroups

\begin{equation*} U_n \leq U \leq C^\times\text{.} \end{equation*}

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Example 6.5. Special Linear Group.

Recall the general linear group, \(\operatorname{GL}_n(\F)\text{,}\) from Example 2.25 is defined to be the set of \(n \times n\) matrices with non-zero determinant. Contained within this group are the matrices with determinant one, called the Special Linear Group,

\begin{equation*} \operatorname{SL}_n(\F) = \left\{M \in \operatorname{GL}_n(\F) \;\middle\vert\; \det{M} = 1\right\}\text{.} \end{equation*}

The set \(\operatorname{SL}_n(\F)\) is closed under multiplication because \(\det{M} = \det{N} = 1\) implies \(\det{MN} = \det{M}\det{N} = 1\text{;}\) contains the identity element because \(\det{I_n} = 1\text{;}\) is closed under inverses because \(\det{M} = 1\) implies \(\det{M^{-1}} = \left(\det{M}\right)^{-1} = 1\text{.}\) Therefore \(\operatorname{SL}_n(\F) \leq \operatorname{GL}_n(\F)\text{.}\)

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Example 6.6. Subgroup of Continuous Real-Valued Functions of a Single Real Variable.

Consider the set of real-valued functions of a single real value, \(\R^{\R}\text{.}\) By Exercise Group 1.3.1–4, this is an additive abelian group under pointwise addition. The subset of continuous functions

\begin{equation*} \mathcal{C}^0(\R,\R) = \left\{f \colon \R \to \R \;\middle\vert\; f\ \text{is continuous}\right\} \subseteq \R^{\R} \end{equation*}

is a subgroup because

The sum of two continuous functions is again continuous,
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The identity for pointwise addition of functions,

\begin{align*} 0 \colon \R \amp\to \R\\ x \amp\mapsto 0\text{,} \end{align*}

is continuous, and

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The inverse of a continuous function \(f\) under pointwise addition of functions, \(-f(x)\text{,}\) is continuous.
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Remark 6.7.

Essentially the same argument, replacing the word "continuous" with "differentiable", proves that the set of differentiable real-valued functions of a single real variable forms a subgroup under pointwise addition of functions.

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We could stipulate further that a function not only has a derivative, but that its derivative is also continuous. Using recursion, we could then define a function, \(f \in \R^\R\text{,}\) to be \(n\)-times continuously differentiable if for all \(k \leq n\text{,}\) the \(k^\text{th}\) derivative, \(f^{(k)}(x)\text{,}\) exists and \(f^{(n)}(x)\) is continuous. We denoted by \(\mathcal{C}^n(\R,\R)\) the set of all \(n\)-times continuously differentiable functions. We say a function is smooth if for all non-negative integers \(n\text{,}\) \(f \in \mathcal{C}^n(\R,\R)\) and denote the set of all such functions by \(\mathcal{C}^\infty(\R,\R)\text{.}\) This construction provides a countably infinite descending chain of subgroups

\begin{equation*} \R^\R \geq \mathcal{C}^0(\R,\R) \geq \mathcal{C}^1(\R,R) \geq \cdots \geq \mathcal{C}^n(\R,\R) \geq \cdots \geq \mathcal{C}^{\infty}(\R,\R)\text{.} \end{equation*}

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Section 6.2 The Subgroup Criterion

Theorem 6.8. The Subgroup Criterion.

Assume \(G\) is a group. For all non-empty \(H \subseteq G\text{,}\) \(H\) is a subgroup if and only if for all \(a,b \in H\text{,}\) \(ab^{-1} \in H\text{.}\)

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Proof.

Let \(H \subseteq G\) be nonempty. Assume \(H\) is a subgroup. By assumption, \(H\) is a group under the restriction of the binary operation to \(H\text{.}\) This means \(H\) is closed under the group operation and closed under inverses. Hence for all \(a,b \in H\text{,}\) \(b^{-1} \in H\) and thus \(ab^{-1} \in H\text{.}\)

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Conversely, assume for all \(a,b \in H\text{,}\) \(ab^{-1} \in H\text{.}\) Since \(H\) is nonempty, there exists an element \(a \in H\text{.}\) Hence \(1 = aa^{-1} \in H\text{.}\) Now we see that \(H\) is closed under inverses because for all \(a \in H\text{,}\)

\begin{equation*} a^{-1} = 1a^{-1} \in H\text{.} \end{equation*}

Finally, we conclude that \(H\) is closed under the binary operation on \(G\) because for all \(a,b \in H\text{,}\) \(b^{-1} \in H\) implies

\begin{equation*} ab = a\left(b^{-1}\right)^{-1} \in H\text{.} \end{equation*}

Therefore \(H\) is a subgroup of \(G\) if and only if for all \(a,b \in H\text{,}\) \(ab^{-1} \in H\text{.}\)

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Section 6.3 Images and Preimages

We can think of a morphism of groups, \(\phi \colon G \to H\text{,}\) as a mechanism to transport subgroups between \(G\) and \(H\) using simple set theoretic constructions.

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Definition 6.9. Image.

Assume \(\phi \colon G \to H\) is a morphism of groups. For all \(K \leq G\text{,}\) the image of \(K\) under \(\phi\) is the set

\begin{equation*} \phi(K) = \left\{\phi(k) \;\middle\vert\; k \in K\right\} \subseteq H\text{.} \end{equation*}

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Lemma 6.10. Morphisms of Groups Preserve the Identity.

If \(\phi \colon G \to H\) is a morphism of groups, then \(\phi(1_G) = 1_H\)

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Proof.

This is a direct consequence of Definition 3.2 and Right Cancellation:

\begin{align*} 1_H\phi(1_G) \amp= \phi(1_G)\\ \amp= \phi(1_G 1_G)\\ \amp= \phi(1_G)\phi(1_G) \end{align*}

implies \(\phi(1_G) = 1_H\text{.}\)

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Lemma 6.11. Morphisms of Groups Preserve Inverses.

If \(\phi \colon G \to H\) is a morphism of groups, then for all \(x \in G\text{,}\) \(\phi\left(x^{-1}\right) = \phi(x)^{-1}\text{.}\)

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Proof.

Let \(x \in G\) be given. Observe that

\begin{align*} 1_H \amp= \phi(1_G)\\ \amp= \phi\left(xx^{-1}\right)\\ \amp= \phi(x)\phi\left(x^{-1}\right) \end{align*}

implies \(\phi\left(x^{-1}\right)\) and \(\phi(x)^{-1}\) are both solutions to the equation \(\phi(x)y = 1_H\text{.}\) Therefore \(\phi\left(x^{-1}\right) = \phi(x)^{-1}\) by Corollary 2.32.

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Proposition 6.12. Images are Subgroups of the Codomain.

Assume \(\phi \colon G \to H\) is a morphism of groups. For all \(K \leq G\text{,}\) \(\phi(K) \leq H\)

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Proof.

Let \(K \leq G\) be given. First observe that the image is nonempty because \(1_H = \phi(1_G) \in \phi(K)\) by Lemma 6.10.

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Let \(y_1, y_2 \in \phi(K)\) be given. There exist \(x_1, x_2 \in K\) such that \(y_1 = \phi(x_1)\) and \(y_2 = \phi(x_2)\) by Definition 6.9. By Lemma 6.11,

\begin{equation*} y_1y_2^{-1} = \phi(x_1)\phi(x_2)^{-1} = \phi(x_1)\phi\left(x_2^{-1}\right) = \phi\left(x_1x_2^{-1}\right) \in \phi(K)\text{.} \end{equation*}

Therefore \(\phi(K)\) is a subgroup by The Subgroup Criterion.

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Definition 6.13. Preimage.

Assume \(\phi \colon G \to H\) is a morphism of groups. For all \(L \leq H\text{,}\) the preimage of \(L\) under \(\phi\) is the set

\begin{equation*} \phi^{-1}(L) = \left\{\phi(\ell) \;\middle\vert\; \ell \in L\right\} \subseteq G\text{.} \end{equation*}

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Proposition 6.14. Preimages are Subgroups of the Domain.

Assume \(\phi \colon G \to H\) is a morphism of groups. For all \(L \leq H\text{,}\) \(\phi^{-1}(L) \leq G\)

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Proof.

Assume \(\phi \colon G \to H\) is a morphism of groups and \(L \leq H\text{.}\) First observe that \(\phi^{-1}(L)\) is nonempty because \(\phi(1_G) = 1_H \in L\) implies \(1_G \in \phi^{-1}(L)\text{.}\)

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Let \(x_1, x_2 \in \phi^{-1}(L)\) be given. Since \(\phi(x_1), \phi(x_2) \in L\text{,}\) it follows that

\begin{equation*} \phi(x_1x_2^{-1}) = \phi(x_1)\phi(x_2)^{-1} \in L \end{equation*}

and thus \(x_1x_2^{-1} \in \phi^{-1}(L)\text{.}\) Therefore \(\phi^{-1}(L) \leq G\) by The Subgroup Criterion.

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Section 6.4 Kernels

Definition 6.15. Kernel of a Morphism of Groups.

Assume \(\phi \colon G \to H\) is a morphism of groups. The preimage of the trivial subgroup in \(H\) is called the kernel subgroup of \(\phi\)

\begin{equation*} \ker{\phi} = \phi^{-1}(\{1_H\}) = \left\{x \in G \;\middle\vert\; \phi(x) = 1_H\right\}\text{.} \end{equation*}

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Remark 6.16.

Note that Proposition 6.14 implies \(\ker{\phi} \leq G\text{.}\)

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Theorem 6.17.

Assume \(\phi \colon G \to H\) is a morphism of groups. Then \(\phi\) is injective if and only if \(\ker{\phi} = \{1_G\}\text{.}\)

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Proof.

We leave the proof as an exercise for the reader.

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Theorem 6.18. Universal Property of Kernels.

Assume \(\phi \colon G \to H\) is a morphism of groups and \(K = \ker{\phi}\text{.}\) Let \(\iota \colon K \to G\) be the inclusion of \(K\) into \(G\text{.}\) For every morphism of groups \(\psi \colon L \to G\text{,}\) if for all \(\ell \in L\text{,}\) \(\phi \circ \psi (\ell) = 1\text{,}\) then there exists a unique morphism of groups \(\overline{\psi} \colon L \to K\) such that

\begin{equation*} \psi = \iota \circ \overline{\psi}\text{.} \end{equation*}

In this case, we say that \(\psi\) factors through \(\iota\).

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We represent this universal property via the diagram below, where \(1 : L \to H\) is defined by the property for all \(\ell \in L\text{,}\) \(\ell \mapsto 1 \in H\text{.}\)

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Figure 6.19. The diagram representing the universal property for the kernel.
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Proof.

Assume \(\phi \colon G \to H\) is a morphism of groups and \(K = \ker{\phi}\text{.}\) Let \(\iota \colon K \to G\) be the inclusion of \(K\) into \(G\text{.}\) Suppose \(\psi \colon L \to G\) is a morphism of groups and for all \(\ell \in L\text{,}\) \(\phi \circ \psi(\ell) = 1\text{.}\) By definition, for every \(\ell \in L\text{,}\) \(\psi(\ell) \in K\text{.}\) Hence we define

\begin{align*} \overline{\psi} \colon L \amp\to K\\ x \amp\mapsto \psi(x) \end{align*}

and observe that for all \(\ell \in L\text{,}\)

\begin{equation*} \iota \circ \overline{\psi}(\ell) = \iota \circ \psi(\ell) = \psi(\ell)\text{.} \end{equation*}

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If \(\omega \colon L to K\) is another morphism of groups with the property that \(\iota \circ \omega = \psi\text{,}\) then for all \(\ell \in L\text{,}\)

\begin{equation*} \omega(\ell) = \iota \circ \omega(\ell) = \psi(\ell) = \overline{\psi}(\ell)\text{.} \end{equation*}

Hence \(\omega = \overline{\psi}\text{.}\) Therefore \(\overline{\psi}\) is the unique morphism \(L \to K\) with this property.

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Exercises 6.5 Exercises for Undergrads & Grads

1.

Prove Theorem 6.17

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Hint.

Recall \(\phi \colon G \to H\) is injective means for all \(a,b \in G\text{,}\) if \(\phi(a) = \phi(b)\text{,}\) then \(a = b\text{.}\) Use Lemma 6.10 to prove the forward direction. For the reverse direction, start by showing that if \(\phi(a) = \phi(b)\text{,}\) then \(\phi(ab^{-1}) \in \ker{\phi}\text{.}\)

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Solution.

Assume \(\phi\) is injective. By Lemma 6.10, \(\phi(1_G) = 1_H\) implies \(1_G \in \ker{\phi}\text{.}\) Hence \(\{1_G\} \leq \ker{\phi}\text{.}\) For the reverse inclusion, observe that

\begin{equation*} g \in \ker{\phi} \iff \phi(g) = 1 = \phi(1) \implies g = 1 \end{equation*}

because \(\phi\) is injective. This establishes \(\ker{\phi} \leq \{1_G\}\text{.}\) Since \(\leq\) is anti-symmetric, we conclude that \(\ker{\phi} = \{1_G\}\text{.}\) Hence if \(\phi\) is injective, then \(\ker{\phi} = \{1_G\}\text{.}\)

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Conversely, assume \(\ker{\phi} = \{1_G\}\text{.}\) Let \(a,b \in G\) be given and suppose \(\phi(a) = \phi(b)\text{.}\) Then

\begin{equation*} 1_H = \phi(a)\phi(a)^{-1} = \phi(a)\phi(b)^{-1} = \phi\left(ab^{-1}\right) \end{equation*}

implies \(ab^{-1} \in \ker{\phi} = \{1_G\}\) and thus \(ab^{-1} = 1_G\text{.}\) Since \(a\) and \(b\) are both solutions to the equation \(yb^{-1} = 1\text{,}\) we conclude \(a = b\) by Corollary 2.32. Hence if \(\ker{\phi} = \{1_G\}\text{,}\) then \(\phi\) is injective. Therefore \(\phi\) is injective if and only if \(\ker{\phi}\) is trivial.

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2. Intersections of Subgroups are Subgroups.

Assume \(G\) is a group and \(H,K \leq G\text{.}\) Prove that \(H \cap K \leq G\text{.}\)

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Solution.

First observe that \(1 \in H\) and \(1 \in K\) imply \(1 \in H \cap K\text{.}\) Let \(x,y \in H \cap K\) be given. Since \(H\) and \(K\) are both subgroups, \(xy^{-1} \in H\) and \(xy^{-1} \in K\) by The Subgroup Criterion Hence \(xy^{-1} \in H \cap K\text{.}\) Therefore \(H \cap K \leq G\) by The Subgroup Criterion.

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3. Arbitrary Intersections of Subgroups are Subgroups.

Assume \(G\) is a group, \(I\) is a set, and \(\{H_i\}_{i \in I}\) is a family of subgroups—that is to say, for every \(i \in I\text{,}\) \(H_i \leq G\text{.}\) Recall that

\begin{equation*} \bigcap_{i \in I} H_i = \left\{x \in G \;\middle\vert\; \forall i \in I, x \in H_i\right\}\text{.} \end{equation*}

or, equivalently,

\begin{equation*} \forall x \in G,\, x \in \bigcap_{i \in I} H_i \iff \forall i \in I, x \in H_i\text{.} \end{equation*}

Prove that \(\bigcap_{i \in I} H_i \leq G\text{.}\)

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Note: No assumptions are made on the set \(I\text{.}\) In particular, \(I\) may be uncountably infinite, so induction is not a valid method of proof here.

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Solution.

First observe that for all \(i \in I\text{,}\) \(1 \in H_i\) implies \(1 \in \bigcap_{i \in I} H_i\text{.}\) Let \(x, y \in G \in \bigcap_{i \in I} H_i\) be given. For every \(i \in I\text{,}\) \(x,y \in H_i\) implies

\begin{equation*} xy^{-1} \in H_i \end{equation*}

by The Subgroup Criterion. Hence \(xy^{-1} \in \bigcap_{i \in I} H_i\text{.}\) Therefore \(\bigcap_{i \in I} H_i \leq G\) by The Subgroup Criterion.

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