Suppose, towards a contradiction, that there is a group isomorphism
\[\phi:(\Q, +) \to (\Q_{ > 0}, \times).\]

Then since $\phi$ is in particular surjective, there exists $r\in \Q$ such that $\phi(r)=2$.
As $r$ is a rational number, so is $r/2$.

It follows that we have
\begin{align*}
2&=\phi(r)=\phi\left(\, \frac{r}{2}+\frac{r}{2} \,\right)\\
&=\phi\left(\, \frac{r}{2} \,\right)\cdot\phi\left(\, \frac{r}{2} \,\right) &&\text{ because $\phi$ is a homomorphism}\\
&=\phi\left(\, \frac{r}{2} \,\right)^2.
\end{align*}

It yields that
\[\phi\left(\, \frac{r}{2} \,\right)=\pm \sqrt{2}.\]

However, this is a contradiction since $\phi\left(\, \frac{r}{2} \,\right)$ must be a positive rational number, yet $\sqrt{2}$ is not a rational number.

We conclude that there is no such group isomorphism, and hence the groups $(\Q, +)$ and $(\Q_{ > 0}\times)$ are not isomorphic as groups.

Multiplicative Groups of Real Numbers and Complex Numbers are not Isomorphic
Let $\R^{\times}=\R\setminus \{0\}$ be the multiplicative group of real numbers.
Let $\C^{\times}=\C\setminus \{0\}$ be the multiplicative group of complex numbers.
Then show that $\R^{\times}$ and $\C^{\times}$ are not isomorphic as groups.
Recall.
Let $G$ and $K$ […]

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 […]

The Set of Square Elements in the Multiplicative Group $(\Zmod{p})^*$
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$.
(a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$.
(b) Determine the index $[G : S]$.
(c) Assume […]

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 […]

Normal Subgroups, Isomorphic Quotients, But Not Isomorphic
Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$.
Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.
Proof.
We give a […]