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I am following JB Fraleigh: a first course in abstract algebra 7th edition. On page 108 he states the Fundamental theorem of finitely generated abelian groups. On page 109 he finds all abelian groups up to isomorphism of order 360. I don't understand this, because I have no way of knowing that all abelian groups of order 360 are finitely generated:

  1. Are all groups generated by some set? I take it the answer is yes: If G is a group, then $\langle G \rangle = G$. Is this correct?
  2. If it is false, why can Fraleigh conclude that all abelian groups of order $n$ are finitely generated?
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    $\begingroup$ Any finite group is finitely generated (just use all the elements as generators). $\endgroup$ – lulu Feb 24 '18 at 16:54
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    $\begingroup$ The answer to your first question yes, so in particular, a finite group is finitely generated. $\endgroup$ – saulspatz Feb 24 '18 at 16:56
  • $\begingroup$ Well... the answer to the first question is all correct if you consider the elements as belonging to $G$ (in other words, $G = \langle G \rangle$ as a subgroup of $G$). But if you only take one symbol for each element of $G$ and build the group $\langle G \rangle$, then this is the free group on the set $G$ of generators, which in general is much bigger than the group $G$. Still, $G$ is a quotient of it, so yes, it is finitely generated if it is finite. All this might look pedantic, but I just wanted to point out that, in general, not only the generators, but also the relations are important. $\endgroup$ – 57Jimmy Feb 24 '18 at 17:01
  • $\begingroup$ Every group of order $360$ is generated by the whole set of its $360$ members. That set of generators is not minimal; that's a more complicated question. $\endgroup$ – Michael Hardy Feb 24 '18 at 17:24
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    $\begingroup$ @57Jimmy: Conceptually what you write is not correct. From the context, the topic of discussion is subsets of a group $G$. Given a group $G$ and a subset $S \subset G$, it makes perfect sense to ask whether $S$ generates $G$: by definition this means that the intersection of all subgroups of $G$ containing $S$ equals $G$. $\endgroup$ – Lee Mosher Feb 24 '18 at 17:42
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Your (1) is correct. $G = \langle G \rangle$ is always generated by its own elements. If $G$ is finite to begin with, this shows it is finitely generated.

So, (2) is moot. Fraleigh can conclude every abelian group of order 360 is finitely generated (at worst, by the set of its own elements).

In general, to say a group $G$ is generated by a subset $S$ of that group means that every element in $G$ can be written as a finite product of the elements in $S$ and their inverses. If $G$ has a finite subset $S$ with this property, it is said to be finitely generated. If there is no finite subset with this property, it is "not finitely generated" or "infinitely generated." Of course a group can also be finite or not finite. Your argument (1) shows that of the four hypothetical possibilities $\{\text{finitely generated or not}\} \times \{\text{finite or not}\}$, the possibility "finite but not finitely generated" can never occur. All three other possibilities do occur:

(a) Finite + finitely generated. As your argument (1) shows, any finite group is in this category. Note that the group usually has a generating set much smaller than itself. Generating it with all its elements is kind of a "last resort" to guarantee it's finitely generated. For example $S_n$ is generated by the transposition $(12)$ and the long cycle $(123\dots n)$.

(b) An infinite group can be finitely generated. For example $\mathbb{Z}$ (with respect to addition) is generated by $1$.

(c) But an infinite group can also fail to be finitely generated. For example $\mathbb{Q}$ (with respect to addition): any finite list of rational numbers has a least common denominator $d$; then the only numbers that can be written in terms of these (and their inverses) are multiples of $1/d$. So it is not possible to find a finite set of rational numbers that generates all rational numbers.

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