Let $(\Q, +)$ be the additive group of rational numbers and let $(\Q_{ > 0}, \times)$ be the multiplicative group of positive rational numbers.

Prove that $(\Q, +)$ and $(\Q_{ > 0}, \times)$ are not isomorphic as groups.

Proof.

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.

The post The Additive Group of Rational Numbers and The Multiplicative Group of Positive Rational Numbers are Not Isomorphic first appeared on Problems in Mathematics.

Let $\Q=(\Q, +)$ be the additive group of rational numbers.

(a) Prove that every finitely generated subgroup of $(\Q, +)$ is cyclic.

(b) Prove that $\Q$ and $\Q \times \Q$ are not isomorphic as groups.

Proof.

(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 post Every Finitely Generated Subgroup of Additive Group $\Q$ of Rational Numbers is Cyclic first appeared on Problems in Mathematics.

Let $\R=(\R, +)$ be the additive group of real numbers and let $\R^{\times}=(\R\setminus\{0\}, \cdot)$ be the multiplicative group of real numbers.

(a) Prove that the map $\exp:\R \to \R^{\times}$ defined by
\[\exp(x)=e^x\] is an injective group homomorphism.

(b) Prove that the additive group $\R$ is isomorphic to the multiplicative group
\[\R^{+}=\{x \in \R \mid x > 0\}.\]

Proof.

(a) Prove $\exp:\R \to \R^{\times}$ is an injective group homomorphism.

We first prove that $\exp$ is a group homomorphism.
Let $x, y \in \R$. Then we have
\begin{align*}
\exp(x+y)&=e^{x+y}\\
&=e^x e^y\\
&=\exp(x)\exp(y).
\end{align*}
Thus, the map $\exp$ is a group homomorphism.

To show that $\exp$ is injective, suppose $\exp(x)=\exp(y)$ for $x, y\in \R$.
This implies that we have
\[e^{x}=e^{y},\] and thus $x=y$ by taking $\log$ of both sides.
Hence $\exp$ is an injective group homomorphism.

(b) Prove that the additive group $\R$ is isomorphic to the multiplicative group $\R^{+}$.

Since the image of $\exp:\R \to \R^{\times}$ consists of positive numbers, we can restrict the codomain of $\exp$ to $\R^{+}$, and we have the injective homomorphism
\[\exp: \R \to \R^{+}.\]

It suffices to show that this homomorphism is surjective.
For any $y\in \R^{+}$, we have $\log(y)\in \R$ and
\[\exp(\log(y))=e^{\log(y)}=y.\]

Thus, $\exp: \R \to \R^{+}$ is a bijective homomorphism, hence isomorphism of groups.
This proves that the additive group $\R$ is isomorphic to the multiplicative group $\R^{+}$.

Note that the inverse homomorphism is given by
\[\log: \R^{+} \to \R\] sending $x\in \R^{+}$ to $\log(x)$.

This is a group homomorphism since we have for $x, y \in \R^{+}$,
\begin{align*}
\log(xy)=\log(x)+\log(y)
\end{align*}
by the property of the log function.

The post The Additive Group $\R$ is Isomorphic to the Multiplicative Group $\R^{+}$ by Exponent Function first appeared on Problems in Mathematics.

Let $F$ be a field and let
\[H(F)=\left\{\, \begin{bmatrix}
1 & a & b \\
0 &1 &c \\
0 & 0 & 1
\end{bmatrix} \quad \middle| \quad \text{ for any} a,b,c\in F\, \right\}\] be the Heisenberg group over $F$.
(The group operation of the Heisenberg group is matrix multiplication.)

Determine which matrices lie in the center of $H(F)$ and prove that the center $Z\big(H(F)\big)$ is isomorphic to the additive group $F$.

Proof.

Suppose that the matrix
\[M=\begin{bmatrix}
1 & x & y \\
0 &1 &z \\
0 & 0 & 1
\end{bmatrix}\] is in the center of the Heisenberg group $H(F)$.
Let
\[A=\begin{bmatrix}
1 & a & b \\
0 &1 &c \\
0 & 0 & 1
\end{bmatrix}\] be an arbitrary element in $H(F)$.

Since $M$ is in the center, we have $AM=MA$, that is,
\[\begin{bmatrix}
1 & a & b \\
0 &1 &c \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & x & y \\
0 &1 &z \\
0 & 0 & 1
\end{bmatrix} =\begin{bmatrix}
1 & x & y \\
0 &1 &z \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & a & b \\
0 &1 &c \\
0 & 0 & 1
\end{bmatrix}.\]

Computing the products, we obtain
\[ \begin{bmatrix}
1 & x+a & y+az+b \\
0 &1 &z+c \\
0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & a+x & b+cx+y \\
0 &1 &c+z \\
0 & 0 & 1
\end{bmatrix}.\] Comparing $(1,3)$-entries, we have
\[az=cx.\] This equality must be true for any $a, c \in F$.

We claim that $x=z=0$.
Taking $a=0, c=1$ (Note that since $F$ is a field, $0, 1 \in F$),
we have $x=0$. Also if $a=1, c=0$, then we have $z=0$.
Thus $x=z=0$ and the matrix $M$ becomes
\[M=\begin{bmatrix}
1 & 0 & y \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix}.\]

It is clear from the computation of $AM=MA$ that this matrix is in the center for any $y$.
Therefore we have determined the center of the Heisenberg group:
\[Z\big(H(F)\big)=\left\{\, \begin{bmatrix}
1 & 0 & y \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix} \quad \middle| \quad \text{ for any } y\in F \, \right \}.\]

To prove that the center $Z\big(H(F)\big)$ is isomorphic to the additive group $F$, consider the map
\[\phi: Z\big(H(F)\big) \to F\] which sends
\[M=\begin{bmatrix}
1 & 0 & y \\
0 &1 & 0 \\
0 & 0 & 1
\end{bmatrix} \in Z\big(H(F)\big)\] to $y\in F$.
We prove that the map $\phi$ is a group isomorphism.

Let
\[M=\begin{bmatrix}
1 & 0 & y \\
0 &1 & 0 \\
0 & 0 & 1
\end{bmatrix}, M’=\begin{bmatrix}
1 & 0 & y’ \\
0 &1 & 0 \\
0 & 0 & 1
\end{bmatrix}\] be any two elements in the center $Z\big(H(F)\big)$.
Then we have
\[MM’=\begin{bmatrix}
1 & 0 & y+y’ \\
0 &1 & 0 \\
0 & 0 & 0
\end{bmatrix}.\] Therefore we have
\[\phi(MM’)=y+y’=\phi(M)+\phi(M’).\] Thus, $\phi$ is a group homomorphism.

From the definition of $\phi$, it is easy to see that the homomorphism $\phi$ is injective and surjective, and hence $\phi$ is a group isomorphism.
Therefore, the center $Z\big(H(F)\big)$ of the Heisenberg group is isomorphic to the additive group of $F$.

Related Question.

The inverse element of the matrix
\[\begin{bmatrix}
1 & x & y \\
0 &1 &z \\
0 & 0 & 1
\end{bmatrix}\] is given by
\[\begin{bmatrix}
1 & -x & xz-y \\
0 & 1 & -z \\
0 & 0 & 1
\end{bmatrix}.\] For a proof, see the post The inverse matrix of an upper triangular matrix with variables.

The post The Center of the Heisenberg Group Over a Field $F$ is Isomorphic to the Additive Group $F$ first appeared on Problems in Mathematics.

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 $K$ be groups.
Recall that a map $f: G \to K$ is a group homomorphism if
\[f(ab)=f(a)f(b)\] for all $a, b \in G$.
A group isomorphism is a bijective homomorphism.
If there is a group isomorphism from $G$ to $K$, we say that $G$ and $K$ are isomorphic (as groups).
 
We give two proofs.

Proof 1.

Seeking a contradiction, assume that there is a group isomorphism $\phi: \C^{\times} \to \R^{\times}$.
Since $\phi$ is a group homomorphism, $\phi(1)=1$. Thus we have
\[1=\phi(1)=\phi((-1)(-1))=\phi(-1)\phi(-1)=\phi(-1)^2.\] Hence $\phi(-1)=\pm 1$. But since $\phi$ is injective and $\phi(1)=1$, we must have $\phi(-1)=-1$.

Now we have
\begin{align*}
-1=\phi(-1)=\phi(i^2)=\phi(i)^2.
\end{align*}

Since $\phi(i)\in \R^{\times}$, $\phi(i)^2$ must be a positive number. Thus we reached a contradiction.
Hence there is no isomorphism between $\R^{\times}$ and $\C^{\times}$.

Proof 2.

Suppose that there is a group isomorphism $\phi: \C^{\times} \to \R^{\times}$.
We want to find a contradiction.
Let $\zeta=e^{2\pi i /3}$ be a primitive third root of unity.

Since $\zeta^3=1$ and $\phi$ is a group homomorphism, we have
\begin{align*}
1=\phi(1)=\phi(\zeta^3)=\phi(\zeta)^3.
\end{align*}
Since $\phi(\zeta)$ is a real number, this implies that $\phi(\zeta)=1$.

This is a contradiction since $\phi$ is injective, but we have $\phi(1)=1=\phi(\zeta)$.
Therefore, there cannot be a group isomorphism between $\R^{\times}$ and $\C^{\times}$.

The post Multiplicative Groups of Real Numbers and Complex Numbers are not Isomorphic first appeared on Problems in Mathematics.