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# Section11.5Subsystems¶ permalink

The subsystem is a fundamental concept of algebra at the universal level.

##### Definition11.5.1Subsystem

If $\left[V; *_1, \ldots ,*_n\right]$ is an algebraic system of a certain kind and $W$ is a subset of $V$, then $W$ is a subsystem of $V$ if $\left[W; *_1, \ldots ,*_n\right]$ is an algebraic system of the same kind as $V$. The usual notation for “$W$ is a subsystem of $V$” is $W \leq V$.

Since the definition of a subsystem is at the universal level, we can cite examples of the concept of subsystems at both the axiomatic and concrete level.

##### Example11.5.2Examples of Subsystems

1. (Axiomatic) If $[G; *]$ is a group, and $H$ is a subset of $G\text{,}$ then $H$ is a subgroup of $G$ if $[H; *]$ is a group.

2. (Concrete) $U = \{-1,1\}$ is a subgroup of $\left[\mathbb{R}^*;\cdot \right]$. Take the time now to write out the multiplication table of $U$ and convince yourself that $[U;\cdot ]$ is a group.

3. (Concrete) The even integers, $2\mathbb{Z} = \{2k : k \textrm{ is} \textrm{ an} \textrm{ integer}\}$ is a subgroup of $[\mathbb{Z}; +]$. Convince yourself of this fact.

4. (Concrete) The set of nonnegative integers is not a subgroup of $[\mathbb{Z}; +]$. All of the group axioms are true for this subset except one: no positive integer has a positive additive inverse. Therefore, the inverse property is not true. Note that every group axiom must be true for a subset to be a subgroup.

5. (Axiomatic) If $M$ is a monoid and $P$ is a subset of $M\text{,}$ then $P$ is a submonoid of $M$ if $P$ is a monoid.

6. (Concrete) If $B^*$ is the set of strings of 0's and 1's of length zero or more with the operation of concatenation, then two examples of submonoids of $B^*$ are: (i) the set of strings of even length, and (ii) the set of strings that contain no 0's. The set of strings of length less than 50 is not a submonoid because it isn't closed under concatenation. Why isn't the set of strings of length 50 or more a submonoid of $B^*$?

For the remainder of this section, we will concentrate on the properties of subgroups. The first order of business is to establish a systematic way of determining whether a subset of a group is a subgroup.

##### Proof

For every group with at least two elements, there are at least two subgroups: they are the whole group and $\{e\}$. Since these two are automatic, they are not considered very interesting and are called the improper subgroups of the group; $\{e\}$ is sometimes referred to as the trivial subgroup. All other subgroups, if there are any, are called proper subgroups.

We can apply Theorem 11.5.3 at both the concrete and axiomatic levels.

##### Example11.5.4Applying Conditions for a Subgroup

1. (Concrete) We can verify that $2\mathbb{Z} \leq \mathbb{Z}$, as stated in Example 11.5.2. Whenever you want to discuss a subset, you must find some convenient way of describing its elements. An element of $2\mathbb{Z}$ can be described as 2 times an integer; that is, $a \in 2\mathbb{Z}$ is equivalent to $(\exists k)_{\mathbb{Z}}(a = 2k)$. Now we can verify that the three conditions of Theorem 11.5.3 are true for 2$\mathbb{Z}\text{.}$ First, if $a, b \in 2\mathbb{Z}$, then there exist $j, k \in \mathbb{Z}$ such that $a = 2j$ and $b = 2k$. A common error is to write something like $a=2j$ and $b=2j$. This would mean that $a=b$, which is not necessarily true. That is why two different variables are needed to describe $a$ and $b\text{.}$ Returning to our proof, we can add $a$ and $b\text{:}$ $a + b = 2j + 2k = 2(j + k)$. Since $j + k$ is an integer, $a + b$ is an element of $2\mathbb{Z}$. Second, the identity, $0$, belongs to 2$\mathbb{Z}$ ($0 = 2(0)$). Finally, if $a \in 2\mathbb{Z}$ and $a = 2k, -a = -(2k) = 2(-k)$, and $-k\in \mathbb{Z}$, therefore, $-a \in 2\mathbb{Z}$. By Theorem 11.5.3, $2\mathbb{Z} \leq \mathbb{Z}$. How would this argument change if you were asked to prove that $3\mathbb{Z} \leq \mathbb{Z}$? or $n \mathbb{Z} \leq \mathbb{Z}, n \geq 2$?

2. (Concrete) We can prove that $H = \{0, 3, 6, 9\}$ is a subgroup of $\mathbb{Z}_{12}$. First, for each ordered pair $(a, b) \in H \times H$, $a +_{12} b$ is in $H\text{.}$ This can be checked without too much trouble since $\left| H \times H\right| = 16$. Thus we can conclude that $H$ is closed under $+_{12}$. Second, $0\in H$. Third, $-0 = 0$, $-3 = 9$, $-6 = 6$, and $-9 = 3$. Therefore, the inverse of each element of $H$ is in $H\text{.}$

3. (Axiomatic) If $H$ and $K$ are both subgroups of a group $G\text{,}$ then $H \cap K$ is a subgroup of $G\text{.}$ To justify this statement, we have no concrete information to work with, only the facts that $H \leq G$ and $K \leq G$. Our proof that $H \cap K \leq G$ reflects this and is an exercise in applying the definitions of intersection and subgroup, (i) If $a$ and $b$ are elements of $H \cap K$, then $a$ and $b$ both belong to $H\text{,}$ and since $H \leq G$, $a * b$ must be an element of $H\text{.}$ Similarly, $a * b \in K$; therefore, $a * b \in H \cap K$. (ii) The identity of $G$ must belong to both $H$ and $K\text{;}$ hence it belongs to $H \cap K$. (iii) If $a \in H \cap K$, then $a \in H$, and since $H \leq G$, $a^{-1}\in H$. Similarly, $a^{-1}\in K$. Hence, by the theorem, $H \cap K \leq G$. Now that this fact has been established, we can apply it to any pair of subgroups of any group. For example, since $2\mathbb{Z}$ and $3\mathbb{Z}$ are both subgroups of $[\mathbb{Z};+]$, $2\mathbb{Z} \cap 3\mathbb{Z}$ is also a subgroup of $\mathbb{Z}\text{.}$ Note that if $a \in 2\mathbb{Z} \cap 3\mathbb{Z}$, $a$ must have a factor of 3; that is, there exists $k\in \mathbb{Z}$ such that $a = 3k$. In addition, $a$ must be even, therefore $k$ must be even. There exists $j \in \mathbb{Z}$ such that $k = 2j$, therefore $a = 3(2j)= 6j$. This shows that $2\mathbb{Z}\cap 3\mathbb{Z}\subseteq 6\mathbb{Z}$. The opposite containment can easily be established; therefore, $2\mathbb{Z} \cap 3\mathbb{Z} = 6\mathbb{Z}$.

Given a finite group, we can apply Theorem 11.3.15 to obtain a simpler condition for a subset to be a subgroup.

##### Example11.5.6Applying the condition for a subgroup of a finite group

To determine whether $H_1= \{0, 5, 10\}$ and $H_2 = \{0, 4, 8, 12\}$ are subgroups of $\mathbb{Z}_{15}$, we need only write out the addition tables (modulo 15) for these sets. This is easy to do with a bit of Sage code that we include below and then for any modulus and subset, we can generate the body of an addition table. The code is set up for $H_1$ but can be easily adjusted for $H_2\text{.}$

Note that $H_1$ is a subgroup of $\mathbb{Z}_{15}$. Since the interior of the addition table for $H_2$ contains elements that are outside of $H_2$, $H_2$ is not a subgroup of $\mathbb{Z}_{15}$.

One kind of subgroup that merits special mention due to its simplicity is the cyclic subgroup.

##### Definition11.5.7Cyclic Subgroup

If $G$ is a group and $a \in G$, the cyclic subgroup generated by $a\text{,}$ $\langle a \rangle$, is the set of all powers of $a\text{:}$ \begin{equation*} \langle a \rangle = \left\{a^n: n \in \mathbb{Z}\right\}\text{.} \end{equation*} We refer to $a$ as a generator of subgroup $\langle a \rangle$.

A subgroup $H$ of a group $G$ is cyclic if there exists $a \in H$ such that $H = \langle a \rangle$.

##### Definition11.5.8Cyclic Group

A group $G$ is cyclic if there exists $\beta \in G$ such that $\langle \beta \rangle=G$.

##### Note11.5.9

If the operation on $G$ is additive, then $\langle a \rangle = \{(n)a : n \in \mathbb{Z}\}$.

##### Example11.5.10

1. In $[\mathbb{R}^* ; \cdot ]$, $\langle 2 \rangle = \left\{2^n : n \in \mathbb{Z}\right\} = \left\{\ldots ,\frac{1}{16}, \frac{1}{8} ,\frac{1}{4}, \frac{1}{2}, 1, 2, 4, 8, 16,\ldots \right\}$.

2. In $\mathbb{Z}_{15}$, $\langle 6 \rangle = \{0, 3, 6, 9, 12\}$. If $G$ is finite, you need list only the positive powers (or multiples) of $a$ up to the first occurrence of the identity to obtain all of $\langle a \rangle\text{.}$ In $\mathbb{Z}_{15}$ , the multiples of 6 are 6, $(2)6 = 12$, $(3)6=3$, $(4)6=9$, and $(5)6 = 0$. Note that $\{0, 3, 6, 9, 12\}$ is also $\langle 3 \rangle$,$\langle 9 \rangle$, and $\langle 12 \rangle$. This shows that a cyclic subgroup can have different generators.

If you want to list the cyclic subgroups of a group, the following theorem can save you some time.

This is an easy way of seeing, for example, that $\langle 9 \rangle$ in $\mathbb{Z}_{15}$ equals $\langle 6 \rangle$, since $-6 = 9$.

# Subsection11.5.1Exercises for Section 11.5¶ permalink

##### 1

Which of the following subsets of the real numbers is a subgroup of $[\mathbb{R}; +]$?

1. the rational numbers

2. the positive real numbers

3. $\{k/2 \mid k \textrm{ is} \textrm{ an} \textrm{ integer}\}$

4. $\{2^k \mid k \textrm{ is an integer}\}$

5. $\{x \mid -100 \leq x \leq 100\}$

Answer
##### 2

Describe in simpler terms the following subgroups of $\mathbb{Z}\text{:}$

1. $5\mathbb{Z} \cap 4\mathbb{Z}$

2. $4\mathbb{Z} \cap 6\mathbb{Z}$ (be careful)

3. the only finite subgroup of $\mathbb{Z}$

##### 3

Find at least two proper subgroups of $R_3$ , the set of $3\times 3$ rook matrices (see Exercise 11.2.4.5).

Answer
##### 4

Where should you place the following in Figure 11.5.12?

1. $e$

2. $a^{-1}$

3. $x * y$

##### 5

1. List the cyclic subgroups of $\mathbb{Z}_6$ and draw an ordering diagram for the relation “is a subset of” on these subgroups.

2. Do the same for $\mathbb{Z}_{12}$.

3. Do the same for $\mathbb{Z}_8$.

4. On the basis of your results in parts a, b, and c, what would you expect if you did the same with $\mathbb{Z}_{24}$?

Answer
##### 6Subgroups generated by subsets of a group

The concept of a cyclic subgroup is a special case of the concept that we will discuss here. Let $[G; * ]$ be a group and $S$ a nonempty subset of $G\text{.}$ Define the set $\langle S \rangle$ recursively by:

• If $a\in S$, then $a\in \langle S \rangle$.

• If $a, b \in \langle S \rangle$, then $a * b \in \langle S \rangle$, and

• If $a \in \langle S \rangle$, then $a^{-1}\in \langle S \rangle$.

1. By its definition, $\langle S \rangle$ has all of the properties needed to be a subgroup of $G\text{.}$ The only thing that isn't obvious is that the identity of $G$ is in $\langle S \rangle$. Prove that the identity of $G$ is in $\langle S \rangle$.

2. What is $\langle\{9, 15\}\rangle$ in$[\mathbb{Z}; +]$?

3. Prove that if $H \leq G$ and $S \subseteq H$, then $\langle S \rangle\leq H$. This proves that $\langle S \rangle$ is contained in every subgroup of $G$ that contains $S$; that is, $\langle S \rangle =\underset{S\subseteq H, H\leq G}{\cap }H$.

4. Describe $\langle \{0.5, 3\}\rangle$ in $\left[ \mathbb{R}^+;\cdot \right]$ and in $[\mathbb{R}; +]\text{.}$

5. If $j, k \in \mathbb{Z}$, $\langle\{j,k\}\rangle$ is a cyclic subgroup of $\mathbb{Z}$. In terms of $j$ and $k\text{,}$ what is a generator of $\langle \{j, k\}\rangle$?

##### 7

Prove that if $H,K \leq G$, and $H\cup K=G$, then $H = G$ or $K = G$.

Hint Answer