From Dummit & Foote, as usual, $\S$ 2.4 #14.

A group $H$ is called finitely generated if there is a finite set $A$ such that $H = \left \langle A \right \rangle$

(a) Prove that every finite group is finitely generated.
(b) (Prove that $\mathbb{Z}$ is finitely generated - I am comfortable with my proof of this, viz $\left \langle 1, -1 \right \rangle = \mathbb{Z}$ )
(c) Prove that every finitely generated subgroup of the additive group $\mathbb{Q}$ is cyclic [If $H$ is a finitely genereated subgroup of $\mathbb{Q}$, show that $H \leq \left \langle \frac{1}{k} \right \rangle$, where $k$ is the product of all the denominators which appear in a set of generators for $H$

Logically, I'm having a hard time getting started on both (a) and (c). For (a), I knew that $H$ generates itself (if that's the correct way to say it), i.e. $\left \langle G \right \rangle = G$ before looking at this relevant wikipedia page, but can't seem to articulate this in a manner consistent with the (currently inaccessible) definition of

$\left \langle A \right \rangle = \bigcap_{A\subseteq H; H \leq G} H$

or the proven result that this subgroup is the set of all products (... a bunch of elements in $G$ with exponents... "words", they call them)

Does this just follow simply from the above definition?

(c) I know (think?) that to show $H$ cyclic, I must take $h \in H$ and show that $h = a^k$, for some $k \leq |H|$ ... or something like that. But then there's that hint. Where do I begin?

Thanks for your help.

  • 2
    $\begingroup$ $\mathbb{Z} = \langle 1 \rangle$ $\endgroup$
    – jspecter
    Commented Sep 21, 2011 at 2:40
  • 1
    $\begingroup$ @jspecter: Of course! In all the confusion, I forgot that "$-1$" in additive notation is $1^{-1}$. Thanks. $\endgroup$
    – Altar Ego
    Commented Sep 21, 2011 at 2:44

2 Answers 2


For part (a), yes it follows from the definition. Since $G \subseteq G$ and $G \leq G$, we see that $G$ is one of the groups in the intersection, so $\langle G \rangle = \bigcap_{A\subseteq H; H \leq G} H \subseteq G$. But we have that $G \subseteq \langle G \rangle$, also by definition, so $\langle G \rangle = G$.

Hint for part (c): Find the greatest common divisor (over $\mathbb{Q}$) of the finitely many generators.

  • $\begingroup$ Thanks for the clarification on (a). I'll take this hint and run with it... $\endgroup$
    – Altar Ego
    Commented Sep 21, 2011 at 2:45

Addressing the other (equivalent) definition of $\langle G\rangle$: Every element $g$ of $G$ is a product of elements of $G$, namely the product of length one with the single factor $g$. Hence $G\subseteq\langle G\rangle$.

Since $G$ is closed under multiplication and taking inverses, $\langle G\rangle\subseteq G$. It follows that $G=\langle G\rangle$.

Additional hint for part (c): Once you have found the $k$ with $H\subseteq\langle\frac{1}{k}\rangle$, recall that every subgroup of a cyclic group is again cyclic.


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