(a) Prove that every finitely generated subgroup of $(\Q, +)$ is cyclic.
Let $G$ be a finitely generated subgroup of $(\Q, +)$ and let $r_1, \dots, r_n$ be nonzero generators of $G$.
Let us express
\[r_i=\frac{a_i}{b_i},\]
where $a_i, b_i$ are integers.
Let
\[s:=\frac{1}{\prod_{j=1}^n b_j} \in \Q.\]
Then we can write each $r_i$ as
\[r_i=\frac{a_i}{b_i}=\left(\, a_i\prod_{\substack{j=1\\j\neq i}}^n b_i \,\right)\cdot \frac{1}{s}.\]
It follows from the last expressions that the elements $r_i$ is contained in the subgroup $\langle s \rangle$ generated by the element $s$.
Hence $G$ is a subgroup of $\langle s \rangle$.
Since every subgroup of a cyclic group is cyclic, we conclude that $G$ is also cyclic.
(b) Prove that $\Q$ and $\Q \times \Q$ are not isomorphic as groups.
Seeking a contradiction, assume that $\Q$ is isomorphic to the direct product $\Q \times \Q$:
\[\Q\cong \Q\times \Q.\]
Then consider the subgroup $\Z\times \Z$ of $\Q\times \Q$.
We claim that the subgroup $\Z\times \Z$ is not cyclic.
If it were cyclic, then there would be a generator $(a,b)\in \Z\times \Z$.
However, for example, the element $(b, -a)$ cannot be expressed as an integer multiple of $(a, b)$.
To see this, suppose that
\[n(a,b)=(b,-a)\]
for some integer $n$.
Then we have $na=b$ and $nb=-a$. Substituting the first equality into the second one, we obtain
\[n^2a=-a.\]
If $a\neq 0$, then this yields that $n^2=-1$, which is impossible, and hence $a=0$.
Then $na=b$ implies $b=0$ as well.
However, $(a,b)=(0,0)$ is clearly not a generator of $\Z\times \Z$.
Thus we have reached a contradiction and $\Z\times \Z$ is a non-cyclic subgroup of $\Q\times \Q$.
This implies via the isomorphism $\Q\cong \Q \times \Q$ that $\Q$ has a non-cyclic subgroup.
We saw in part (a) that this is impossible.
Therefore, $\Q$ is not isomorphic to $\Q\times \Q$.
The Group of Rational Numbers is Not Finitely Generated
(a) Prove that the additive group $\Q=(\Q, +)$ of rational numbers is not finitely generated.
(b) Prove that the multiplicative group $\Q^*=(\Q\setminus\{0\}, \times)$ of nonzero rational numbers is not finitely generated.
Proof.
(a) Prove that the additive […]
Prove that a Group of Order 217 is Cyclic and Find the Number of Generators
Let $G$ be a finite group of order $217$.
(a) Prove that $G$ is a cyclic group.
(b) Determine the number of generators of the group $G$.
Sylow's Theorem
We will use Sylow's theorem to prove part (a).
For a review of Sylow's theorem, check out the […]
Every Cyclic Group is Abelian
Prove that every cyclic group is abelian.
Proof.
Let $G$ be a cyclic group with a generator $g\in G$.
Namely, we have $G=\langle g \rangle$ (every element in $G$ is some power of $g$.)
Let $a$ and $b$ be arbitrary elements in $G$.
Then there exists […]
A Simple Abelian Group if and only if the Order is a Prime Number
Let $G$ be a group. (Do not assume that $G$ is a finite group.)
Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.
Definition.
A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is […]
A Homomorphism from the Additive Group of Integers to Itself
Let $\Z$ be the additive group of integers. Let $f: \Z \to \Z$ be a group homomorphism.
Then show that there exists an integer $a$ such that
\[f(n)=an\]
for any integer $n$.
Hint.
Let us first recall the definition of a group homomorphism.
A group homomorphism from a […]