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From my teacher's notes: "We say that a subset $S\subseteq G$ generates $G$ is any element $a\in G$ is the result of finite compositions of elements in $S$ or their inverses. A group $G$ is called cyclic if there exists an element $a\in G$ that generates $G$."

I still don't quite understand the definition of "generate", "finite compositions" and "cyclic" after reading the paragraphs. please help.

2 Answers2

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Perhaps a few examples will clarify. Let $S = \{a,b,c\} \subseteq G$.

$abac$ is a finite composition of elements of $S$. Here, "composition" is just another word for the group operation, maybe you also call it "multiplication. Finite means exactly what you think it does. This is a finite composition because it is a composition of finite things. Other examples include $a$, $bbb$, $abcbabbaca$.

We're not just using elements of $S$, we're including "...elements in $S$ or their inverses." So we also allow $\{a^{-1},b^{-1},c^{-1}\}$ in our finite compositions. So now we are allowed for instance $ab^{-1}cb^{-1}abb^{-1}aca^{-1}$.

Thus, $S \subseteq G$ generates $G$ if every element of $G$ can be written in the way just described.

A special case of this is when $S = \{a\}$, in which case $G$ is just all powers of $a$: $G = \{\ldots, a^{-2},a^{-1},e=a^0,a^1,a^2,\ldots\}$ (note that $a$ might have finite order, so $G$ might still be finite.) In this case, we say $G$ is cyclic.

Edit: As mentioned by a comment, "$a$ has finite order" means that there is some $n \in \mathbb{N}$ so that $a^n=e$. In particular, this mean $a^{n+1}=a$. Then $G$ (from above) can be written as

\begin{align} G &= \{\ldots,a^{-2}, a^{-1},e=a^0,a^1,a^2,\ldots, a^{n-1},a^n=e,a^{n+1}=a,\ldots\} \\ &=\{\ldots,a^{-2},a^{-1},e,a,a^2,\ldots,a^{n-2},a^{n-1},e,a,a^2,\ldots \} \end{align}

After $n$ elements, the list repeats, or "cycles." Because sets don't see repitition, we may rewrite as

$$G = \{e,a,a^2,\ldots, a^{n-2},a^{n-1}\}$$

Hence, $G$ is finite.

Noah Caplinger
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  • what does it mean for $a$ to have finite order? – timothy5597 Dec 24 '19 at 04:20
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    @timothy5597 It means for some finite $n, a^n=1.$ For example, in the additive group $\Bbb Z/34 \Bbb Z$, the element $1$ has order $34,$ but in the additive group $\Bbb Z$, the element $1$ has infinite order. – Robert Shore Dec 24 '19 at 07:19
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As explained in another answer, a subset $S$ of a group $G$ generates $G$ if every element of $G$ is a finite product of elements of $S$ or their inverses. Let me elaborate on why that is the definition, since this idea of "generation" appears in many other contexts.

If $S \subseteq G$, we can define the subgroup of $G$ generated by $S$, denoted $\langle S \rangle$; once we define this, then $G$ is generated by $S$ precisely if the subgroup of $G$ generated by $S$ is in fact all of $G$ (i.e. if $\langle S \rangle = G$). So how do we define the subgroup of $G$ generated by $S$? It's the smallest subgroup of $G$ which contains every element of $S$. That is, it contains $S$, along with the minimum amount of other stuff we need to include in order to have a subgroup of $G$. It is a subgroup of $G$ containing $S$, and there is no other subgroup $H$ such that $S \subseteq H \trianglelefteq \langle S \rangle$.

So what is the "minimum amount of other stuff" we need to add to $S$ in order to make it into a subgroup of $G$? Well, subgroups are closed under taking inverses, so had better include the inverse of every element of $S$. Subgroups are also closed under multiplication (i.e. the group operation of $G$), so along with elements of $S$ and their inverses, we also need to include all finite products of elements of $S$ and their inverses. This is enough though; we have obtained a subgroup. Notice that it's closed under multiplication; using the example $S = \{a, b, c\}$, if $abcca^{-1}b$ and $cb^{-1}a^{-1}b$ are two finite products of elements of $S$ and their inverses, their product $abcca^{-1}bcb^{-1}a^{-1}b$ is also a finite product of elements of $S$ and their inverses! It's also closed under taking inverses; for example, the inverse of $cb^{-1}a^{-1}b$ is $b^{-1}abc^{-1}$ (which again is a finite product of elements of $S$ and their inverses). In summary, we have produced $\langle S \rangle$ by adding to $S$ only the things we need to make it a subgroup.

Lastly, let me briefly describe how this idea generalizes to other contexts. In linear algebra, the subspace $\langle S \rangle$ of a vector space $V$ spanned by $S$ of vectors is the smallest subspace of $V$ containing $S$. The details are different here, since vector spaces are different than groups ($\langle S \rangle$ comprises all linear combinations of finite subsets of $S$), but the core idea is the same. In ring theory, the ideal generated by a subset of a ring is the smallest ideal containing the subset. In topology (or in the study of metric spaces), the closure of a set $A$ is the smallest closed set containing $A$; it is, if you like, the "closed set generated by $A$."

  • does a linear span of the base vectors in a vector space coincide with $\langle S\rangle$ in a group? I ask just in case I've missed something. Then there are many parallels with the quotient space? – PinkyWay Dec 24 '19 at 13:04
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    They are analogous concepts. The subgroup of $G$ generated by $S$ is the smallest subgroup of $G$ containing $S$; the vector subspace of $V$ spanned by $S$ is the smallest subspace of $V$ containing $S$. (In the vector space case, $S$ can be a set of basis vectors, but doesn't have to be.) – Justin Barhite Dec 24 '19 at 13:15