Generators and isomophisms

Let $G$ be an arbitrary group, let $S$ be a complex of $G$, and let $\{H_\alpha\}_{\alpha\in\Lambda}$ be the collection of all subgroups of $G$ which contain $S$, i.e., $S\subset H_\alpha$ for all $\alpha\in\Lambda$. The collection, $\{H_\alpha\}_{\alpha\in\Lambda}$, is clearly not empty since $G$ itself is a subgroup which contains $S$. We denote the intersection of all subgroups of $G$ containing $S$ by $gp_G(S)$. That is

$$gp_G(S)=\cap_{\alpha\in\Lambda}H_\alpha,$$

or

$$gp_G(S)=\cap_{S\subset H\leq G}H.$$

(When it is clear that $G$ is the group being considered, we use $gp(S)$.) From exercise 6 for Chapter 2, we have $gp(S)\leq G$. From its very definition, we have $S\subset gp(S)$ and $gp(S)$ is contained in any other subgroup which contains $S$. In this sense, $gp_G(S)$ is the smallest subgroup of $G$ containing $S$. We call $gp(S)$ the group generated by $S$.

Proposition 5.1.1   Let $G$ be a group, $S$ a complex of $G$, and let $E$ be the set of all finite products of elements of $S$ and their inverses (including single elements of $S$). Then $gp(S) = E$.

Proof: It is readily seen that $E$ is a subgroup of $G$ (see exercise 1 for this section) and $S\subset E$. Thus $gp(S) \subset E$. However, since $gp(S)\leq G$ and $S\subset gp(S)$, $gp(S)$ must certainly contain all finite products of elements of $S$ and of inverses of elements of $S$, i.e., $E \subset gp(S)$. Thus $E = gp(S)$. $\Box$

Let us consider some examples.

Example 5.1.2   Proposition 5.1.1 shows that in $S_4$, $gp(\{(1,2,4), (2,3,4)\}) = gp(\{(1,2,3), (1,2) (3,4)\})$. To see this, we note that

$$(1,2,4)=(1,2,3)(1,2)(3,4)(1,2,3),$$

$$(2,3,4)=(1,3,2)(1,2)(3,4)=(1,2,3)^{-1}(1,2)(3,4),$$

which shows $gp(\{(1,2,4), (2,3,4)\})\subset gp(\{(1,2,3), (1,2) (3,4)\})$. To get the reverse inclusion, we note that

$$(1,2,3)=(1,2,4)(2,3,4),$$

$$(1,2)(3,4)=(2,3,4)(1,4,2)=(2,3,4)(1,4,2)^{-1}.$$

$\Box$

We now turn our attention to the important case for this chapter. If $S = \{a\}$, then we shall write $\langle a\rangle$ for $gp(\{a\})$; and $\langle a\rangle$ will be referred to as the cyclic subgroup generated by $a$. The group G itself is called cyclic (or a cyclic group) if there exists an $a \in G$ such that $G = \langle a\rangle$, and such an element $a$ is called a generator of $G$.

Example 5.1.3   The set of integers $\mathbb {Z}$ under ordinary addition is cyclic generated by $1$. (Note that $- 1$ is also a generator. Also recall that when the operation is addition, $1^n$ is interpreted as $n\cdot 1=1+1+...+1$ ($n$ times), when $n >0$, and as $n\cdot 1=(-1)+(-1)+...+(-1)$ ($\vert n\vert$ times), when $n < 0$.) $\Box$

Example 5.1.4   If $G$ is the group of example 2.1.11, then $G$ is cyclic and $G = \langle \zeta_n\rangle$, where $\zeta_n$ is what is called a primitive $n^{th}$ root of unity. If one recalls that $e^{i\theta} = \cos(\theta) + i \sin (\theta)$ (Euler's Formula), then we can take $\zeta_n = e^{2\pi i/n}$. We should note that this example gives us a cyclic group of order $n$ for every positive integer $n$.

This group is the same as that in Example 2.1.11. $\Box$

Before proceeding further in our discussion of cyclic groups, it will be convenient to introduce the notion of isomorphic groups.

Definition 5.1.5   Two groups $G_1$ and $G_2$ are said to be isomorphic if there exists a mapping

$$f:G_1\rightarrow G_2$$

such that

(1)

$f$ is 1-1 and onto,

(2)

$f(ab) = f(a)f(b)$, for all $a,b \in G$.

The second condition is sometimes referred to by saying that ``$f$ preserves the group operation.'' Also it should be noted, we have designated the operation in a multiplicative fashion (or juxtapositive) in both groups; although we warn the reader that the elements of $G_1$ and $G_2$ might be of an entirely different nature, as well as the operations defined between them. Nevertheless from an abstract point of view, isomorphic groups are indistinguishable. In other words, if $G_1$ and $G_2$ are isomorphic, then any relationship involving the binary operation holding for elements of one of the $G_i$ ($i = 1, 2$) holds for the corresponding elements under the mapping for the other $G_i$ ($i = 1, 2$). For example, suppose $f : G_1 \rightarrow G_2$ satisfies the conditions of Definition 5.1.5, i.e., $G_1$ and $G_2$ are isomorphic. Then if $G_1$ is abelian, $G_2$ must be abelian. If $e_1$ is the identity of $G_1$, then $f(e_1)= e_2$ is the identity of $G_2$ (see exercise 3 for this section). If $a^n = e_1$ in $G_1$, then $f(a)^n = f(a^n) = f(e_1) = e_2$, etc. Thus although their elements might be quite different, $G_1$ and $G_2$ are abstractly indistinguishable.

A mapping $f$ satisfying the conditions of Definition 5.1.5 is called an isomorphism of $G_1$ onto $G_2$. If $G_1$ and $G_2$ are isomorphic, we will write $G_1 \cong G_2$. What we have noted above in words is that an isomorphism takes an identity into an identity, an element of order $n$ into an element of order $n$ (this is somewhat stronger than just saying $a^n = e_1$ implies $f(a)^n = e_2$. Why?), etc. We also note that if we have the Cayley table for $G_1$, then we can use f to write the Cayley table for $G_2$ since $f$ preserves the operation.

Example 5.1.6   Consider $\mathbb {C}$ the additive group of all complex numbers and the subgroup $\hat{\mathbb {C}}$ of the group $\mathbb {C}^n$ of example 2.1.14 of all $n$-tuples of the form $(\alpha, 0, ..., 0)$, $\alpha\in \mathbb {C}$. Clearly the mapping

$$\begin{array}{ccc} \mathbb {C}& \rightarrow & \mathbb {C}\\ \alpha & \longmapsto & (\alpha,0,...,0). \end{array}$$

is an isomorphism of $\mathbb {C}$ onto $\hat{\mathbb {C}}$, or $\mathbb {C}\cong \hat{\mathbb {C}}$. (Verify this for yourself!) $\Box$