For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$
\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]

For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by
\[ (f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S . \]

(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?

(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.

(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.

(d) Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as
\[ V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} . \]

(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?

(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.

Proof.

(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?

We will prove that each vector space axioms holds for $\Fun(S, V)$.

Closed under addition and scalar multiplication: For $f , g \in \Fun(S, V)$ and $c, d \in \K$, $cf+dg$ is another function with domain $S$ defined by
\[ (cf+dg)(s) = cf(s) + dg(s) \in V \, \mbox{ for all } s \in S . \] Thus $cf+dg \in \Fun(S, V)$. This proves that $\Fun(S, V)$ is closed under addition and scalar multiplication.

Addition is associative: For $f, g, h \in \Fun(S, V)$ and $s \in S$ we have
\begin{align*}
((f+g)+h)(s) &= (f+g)(s) + h(s) \\
&= ( f(s) + g(s) ) + h(s) \\
&= f(s) + ( g(s) + h(s) ) \\
&= f(s) + (g+h)(s) \\
&= (f+(g+h))(s) . \end{align*}

Because $((f+g)+h)(s) = (f+(g+h))(s)$ holds for each $s \in S$, we can say that $(f+g)+h = f+(g+h)$.

Addition is commutative: Notice that for $s \in S$ we have
\[ (f+g)(s) = f(s) + g(s) = g(s) + f(s) = (g+f)(s) . \] Because this holds for every $s \in S$, we can say that $f+g = g+f$.

Zero vector: The zero vector is the function $\mathbf{0}$, defined by $\mathbf{0}(s) = 0 \in V$ for each $s \in S$. To check that $\mathbf{0}$ is the additive identity, we check for any $f \in \Fun(S, V)$ and $s \in S$,
\[ (f+ \mathbf{0})(s) = f(s) + \mathbf{0}(s) = f(s) + 0 = f(s) . \] Because this holds for each $s \in S$, we can say that $f + \mathbf{0} = f$.

Inverse vectors: For $f \in \Fun(S, V)$, its additive inverse is the function $(-f)$ defined by
\[ (-f)(s) = – f(s) \, \mbox{ for all } s \in S . \] To check that $-f$ is the inverse of $f$, we check for all $s \in S$
\[ (f + (-f))(s) = f(s) + (-f)(s) = f(s) – f(s) = 0 = \mathbf{0}(s) . \]

Scalar multiplication is associative: For $c, d \in \K$, $f \in \Fun(S, V)$ and $s \in S$ we have
\[ ((cd)(f))(s) = (cd) f(s) = c ( d f(s) ) = c ( df)(s) = (c (df) )(s) . \] Because this holds for each $s \in S$, we conclude that $(cd)(f) = c (df)$.

Scalar identity element: Let $1 \in \K$ be the multiplicative identity. Then for $f \in \Fun(S, V)$ and $s \in S$ we have
\[ (1f)(s) = 1 f(s) = f(s) . \] Because this holds for all $s \in S$, we conclude that $1f = f$.

Distributivity: Next we check distributivity. For $c, d \in \K$, $f \in \Fun(S, V)$ and $s \in S$, we have
\[ ( (c+d)(f) )(s) = (c+d) f(s) = c f(s) + d f(s) = (cf + df)(s) . \] Because this holds for all $s \in S$, we can conclude that
\[ (c+d)f = cf + df . \]

For $c \in \K$, $f, g \in \Fun(S, V)$ and $s \in S$ we have
\begin{align*}
( c(f+g))(s) &= c ( (f+g)(s) ) \\
&= c ( f(s) + g(s) ) \\
&= cf(s) + cg(s) \\
&= (cf + cg)(s) . \end{align*}

Because this holds for all $s \in S$, we conclude that
\[ c(f+g) = cf + cg . \]

We have shown that $\Fun(S, V)$ satisfies all of the vector space axioms, so it is a vector space.

(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.

Define the map $T : \Fun(S_1 , V) \rightarrow V$ by
\[ T(f) = f(s) \, \mbox{ for all } f \in \Fun(S_1 , V) . \]

We check that $T$ is a linear map. For $f, g \in \Fun(S_1, V)$ we have
\[ T(f+g) = (f+g)(s) = f(s) + g(s) = T(f) + T(g) . \] Thus $T$ is additive. Next, for $c \in \K$ and $f \in \Fun(S_1, V)$ we have
\[ T(cf) = (cf)(s) = c f(s) = c T(f) . \] This shows that $T$ respects scalar multiplication, and so $T$ is a linear map.

Next we prove that $T$ is an isomorphism. To see that $T$ is surjective, pick any $v \in V$. Define the map $f_v \in \Fun(S_1, V)$ by $f_v(s) = v$. Then $T(f_v) = f_v(s) = v$, and so $v$ lies in the range of $T$.

To see that $T$ is injective, it is enough to show that if $T(f) = 0 \in V$ then $f = \mathbf{0} \in \Fun(S_1, V)$. So suppose that $T(f) = 0$. Because $T(f) = f(s)$, this implies that $f(s) = 0$. But $\mathbf{0}(s) = 0$ as well. Because $f(s) = \mathbf{0}(s)$ for each $s \in S_1$ (which, remember, contains only the one element), we can conclude that $f = \mathbf{0}$.

We have shown that $T$ is linear, injective, and surjective. This finishes the proof that $T$ is a linear isomorphism.

(c) Find a basis of $\Fun(S_1 , V)$.

We can use the basis $B$ of $V$ and the linear isomorphism $T$ found in the previous part to define a basis of $\Fun(S_1, V)$. More specifically, the image of $B$ under the inverse isomorphism
\[ T^{-1} : V \rightarrow \Fun(S_1 , V) \] will define a basis of $\Fun(S_1 , V)$.

For $v \in V$ define the function $f_v$ by $f_v(s) = v$. It is clear that $T(f_v) = v$, and so $T^{-1}(v) = f_v$ for all $v \in V$. In particular, $T^{-1}(e_i) = f_{e_i}$ for each $1 \leq i \leq n$, and so the set $\{ f_{e_1} , f_{e_2} , \cdots , f_{e_n} \}$ forms a basis of $\Fun(S, V)$.

(d) Construct a linear isomorphism between $\Fun(S, V)$ and the vector space $V^m$

Define the map $T : \Fun(S, V) \rightarrow V^m$ by
\[ T(f) = ( f(s_1) , f(s_2) , \cdots , f(s_m) ) \in V^m \, \mbox{ for all } f \in \Fun(S, V) . \]

First we show that $T$ is linear. For $f, g \in \Fun(S, V)$ and $c, d \in \K$ we have
\begin{align*}
T(cf+dg) &= \left( (cf + dg)(s_1) , (cf + dg)(s_2) , \cdots , (cf + dg)(s_m) \right) \\
&= \left( (cf)(s_1) + (dg)(s_1) , (cf)(s_2) + (dg)(s_2) , \cdots , (cf)(s_m) + (dg)(s_m) \right) \\
&= \left( c f(s_1) , c f(s_2) , \cdots , c f(s_m) \right) + \left( d g(s_1) , d g(s_2) , \cdots , d g(s_m) \right) \\
&= c \left( f(s_1) , f(s_2) , \cdots , f(s_m) \right) + d \left( g(s_1) , g(s_2) , \cdots , g(s_m) \right) \\
&= c T(f) + d T(g) . \end{align*}

This proves that $T$ is a linear map.

Next we show that it is an isomorphism.

First we show surjectivity. For $(v_1 , v_2 , \cdots , v_m) \in V^m$ define the function $f$ by
\[ f(s_i) = v_i \, \mbox{ for } 1 \leq i \leq m . \]

Then
\[ T(f) = ( f(s_1) , f(s_2) , \cdots , f(s_m) ) = ( v_1 , v_2 , \cdots , v_m ) . \] Thus every element of $V^m$ lies in the range of $T$.

Next we check injectivity. Suppose that $T(f) = (0 , 0 , \cdots , 0) \in V^m$. This means that $f(s_i) = 0$ for each $1 \leq i \leq m$. Thus $f(s) = \mathbf{0}(s)$ for all $s \in S$, and so $f = \mathbf{0}$. This proves that $T$ is injective. Together with surjectivity and linearity, we have proven that $T$ is an isomorphism.

(e) Constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?

We will use the basis $B = \{ e_1 , \cdots , e_n \}$ of $V$ to create a basis for $V^m$. We will translate this basis into a basis of $\Fun(S, V)$ using the isomorphism $T$, or actually its inverse $T^{-1}$.

For integers $i, j$ with $1 \leq i \leq n$ and $1 \leq j \leq m$, define the vector
\[ d_i^j = ( 0 , \cdots , e_i , \cdots , 0 ) . \] Notice that every component is $0$ except for the $j$-th component, which is the vector $e_i$. The set $\{ d_i^j \}_{ \substack{1 \leq i \leq n \\ 1 \leq j \leq m} }$ is the standard basis of $V^m$ defined by $B$. The set $\{ T^{-1} ( d_i^j ) \}_{ \substack{ 1 \leq i \leq n \\ 1 \leq j \leq m}}$ will then be a basis of $\Fun(S, V)$.

To find $T^{-1}$, suppose that $T(f) = d_i^j$. Because the $j$-th component of $d_i^j$ is $e_i$, we must have that $f(s_j) = e_i$. Because all of the other components are $0$, we must have $f(s_k) = 0$ for all $k \neq j$. This information completely determines the function $f$. Define the function $\delta_i^j$ by
\[ \delta_i^j ( s_k ) = \left\{ \begin{array}{cl} e_i & \mbox{ if } k = j \\ 0 & \mbox{ if } k \neq j \end{array} \right. . \]

Then it is clear that $T(\delta_i^j) = d_i^j$; that is, $T^{-1}(d_i^j) = \delta_i^j$. Thus a basis of $\Fun(S, V)$ is formed by the set $\{ \delta_i^j \}_{ \substack{ 1 \leq i \leq n \\ 1 \leq j \leq m }}$. There are exactly $nm$ elements in this basis, thus $\Fun(S, V)$ has dimension $nm$.

(f) Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$.

We will show in one step that $\Fun(S, W)$ is closed under addition and scalar multiplication. For $f, g \in \Fun(S, W)$, $c, d \in \K$ and $s \in S$ we have
\[ (cf + dg)(s) = c f(s) + d g(s) \in W . \] This proves that $(cf + dg) : S \rightarrow W$, and so $(cf + dg) \in \Fun(S, W)$.

The post Vector Space of Functions from a Set to a Vector Space first appeared on Problems in Mathematics.

Let $m$ and $n$ be positive integers such that $m \mid n$.

(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined.

(b) Prove that $\phi$ is a group homomorphism.

(c) Prove that $\phi$ is surjective.

(d) Determine the group structure of the kernel of $\phi$.

Proof.

(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined.

To show that $\phi$ is well-defined, we need show that the value of $\phi$ does not depends on the choice of representative $a$.
So suppose that $a+n\Z=a’+n\Z$ so that $a$ and $a’$ are two representatives for the same element.
This yields that $a-a’$ is divisible by $n$.

Now, $a+n\Z$ is mapped to $a+m\Z$ by $\phi$. On the other hand, $a’+n\Z$ is mapped to $a+m\Z$ by $\phi$.
Since $a-a’$ is divisible by $n$ and $m \mid n$, it follows that $a-a’$ is divisible by $m$.
This implies that $a+m\Z=a’+m\Z$.
This prove that $\phi$ does not depend on the choice of the representative, and hence $\phi$ is well-defined.

(b) Prove that $\phi$ is a group homomorphism.

Let $a+n\Z$, $b+n\Z$ be two elements in $\Zmod{n}$. Then we have
\begin{align*}
&\phi\left(\, (a+n\Z)+(b+n\Z) \,\right)\\
&=\phi\left(\, (a+b)+n\Z) \,\right) &&\text{by addition in $\Zmod{n}$}\\
&=(a+b)+m\Z &&\text{by definition of $\phi$}\\
&=(a+m\Z)+(b+m\Z)&&\text{by addition in $\Zmod{m}$}\\
&=\phi(a+n\Z)+\phi(b+n\Z) &&\text{by definition of $\phi$}.
\end{align*}

Hence $\phi$ is a group homomorphism.

(c) Prove that $\phi$ is surjective.

For any $c+m\Z \in \Zmod{m}$, we pick $c+n\Z\in \Zmod{n}$.
Then as $\phi(c+n\Z)=c+m\Z$, we see that $\phi$ is surjective.

(d) Determine the group structure of the kernel of $\phi$.

If $a+n\Z\in \ker(\phi)$, then we have $0+m\Z=\phi(a+n\Z)=a+m\Z$.
This implies that $m\mid a$.
On the other hand, if $m\mid a$, then $\phi(a+n\Z)=a+m\Z=0+m\Z$ and $a+n\Z\in \ker(\phi)$.

It follows that
\[\ker(\phi)=\{mk+n\Z \mid k=0, 1, \dots, l-1\},\] where $l$ is an integer such that $n=ml$.

Thus, $\ker(\phi)$ is a group of order $l$.
Since $\ker(\phi)$ is a subgroup of the cyclic group $\Zmod{n}$, we know that $\ker(\phi)$ is also cyclic.
Thus
\[\ker(\phi)\cong \Zmod{l}.\]

Another approach

Here is a more direct proof of this result.
Define a map $\psi:\Z\to \ker(\phi)$ by sending $k\in \Z$ to $mk+n\Z$.
It is straightforward to verify that $\psi$ is a surjective group homomorphism and the kernel of $\psi$ is $\ker(\psi)=l\Z$.
It follows from the first isomorphism theorem that
\[\Zmod{l}= \Z/\ker(\psi) \cong \im(\psi)=\ker(\phi). \] Click here if solved 124

The post Group Homomorphism from $\Z/n\Z$ to $\Z/m\Z$ When $m$ Divides $n$ first appeared on Problems in Mathematics.

Let $\R^n$ be an inner product space with inner product $\langle \mathbf{x}, \mathbf{y}\rangle=\mathbf{x}^{\trans}\mathbf{y}$ for $\mathbf{x}, \mathbf{y}\in \R^n$.

A linear transformation $T:\R^n \to \R^n$ is called orthogonal transformation if for all $\mathbf{x}, \mathbf{y}\in \R^n$, it satisfies
\[\langle T(\mathbf{x}), T(\mathbf{y})\rangle=\langle\mathbf{x}, \mathbf{y} \rangle.\]

Prove that if $T:\R^n\to \R^n$ is an orthogonal transformation, then $T$ is an isomorphism.

 

We give two proofs.
The second one uses a fact about the injectivity of linear transformations.

Proof 1.

As $T$ is a linear transformation from $\R^n$ to itself, it suffices to show that $T$ is an injective linear transformation.

Suppose that $T(\mathbf{x})=T(\mathbf{y})$ for $\mathbf{x}, \mathbf{y}\in \R^n$.
We show that $\mathbf{x}=\mathbf{y}$.

We have
\begin{align*}
&\|\mathbf{x}-\mathbf{y}\|^2\\
&=(\mathbf{x}-\mathbf{y})^{\trans}(\mathbf{x}-\mathbf{y})\\
&=(\mathbf{x}^{\trans}-\mathbf{y}^{\trans})(\mathbf{x}-\mathbf{y})\\
&=\mathbf{x}^{\trans}\mathbf{x}-\mathbf{x}^{\trans}\mathbf{y}-\mathbf{y}^{\trans}\mathbf{x}+\mathbf{y}^{\trans}\mathbf{y}\\
&=\langle\mathbf{x}, \mathbf{x} \rangle-\langle\mathbf{x}, \mathbf{y} \rangle-\langle\mathbf{y}, \mathbf{x} \rangle+\langle\mathbf{y}, \mathbf{y} \rangle\\
&=\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{y}) \rangle-\langle T(\mathbf{y}), T(\mathbf{x}) \rangle+\langle T(\mathbf{y}), T(\mathbf{y}) \rangle\\
&\text{(since $T$ is an orthogonal transformation)}\\
&=\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{x}) \rangle-\langle T(\mathbf{x}), T(\mathbf{x}) \rangle+\langle T(\mathbf{x}), T(\mathbf{x}) \rangle\\
&\text{(since $T(\mathbf{x})=T(\mathbf{y})$)}\\
&=0.
\end{align*}
It follows that $\|\mathbf{x}-\mathbf{y}\|=0$ and hence $\mathbf{x}=\mathbf{y}$.
This proves that $T:\R^n\to \R^n$ is injective.

As $T$ is an injective linear transformation from the $n$-dimensional vector space $\R^n$ to itself, it is also surjective, and thus $T$ is an isomorphism.

Proof 2.

Recall that the linear transformation $T$ is injective if and only if the null space $\calN(T)=\{\mathbf{0}\}$, that is, $T(\mathbf{x})=\mathbf{0}$ implies that $\mathbf{x}=\mathbf{0}$.
(See the post “A Linear Transformation is Injective (One-To-One) if and only if the Nullity is Zero” for the proof of this fact.)

We use this fact to show that $T$ is injective.
Suppose that $T(\mathbf{x})=\mathbf{0}$.
Then we have
\begin{align*}
\|\mathbf{x}\|^2&=\langle \mathbf{x}, \mathbf{x}\rangle\\
&=\langle T(\mathbf{x}), T(\mathbf{x})\rangle &&\text{as $T$ is orthogonal}\\
&=\langle \mathbf{0}, \mathbf{0}\rangle=0 &&\text{as $T(\mathbf{x})=\mathbf{0}$}.
\end{align*}

It follows that the length $\|\mathbf{x}\|=0$, and hence $\mathbf{x}=\mathbf{0}$.
This proves that the null space $\calN(T)=\{\mathbf{0}\}$ and $T$ is injective.

As $T$ is an injective linear transformation from $\R^n$ to itself, it is also surjective, and hence $T$ is an isomorphism.

The post An Orthogonal Transformation from $\R^n$ to $\R^n$ is an Isomorphism first appeared on Problems in Mathematics.

Let $N$ be a normal subgroup of a group $G$.
Suppose that $G/N$ is an infinite cyclic group.

Then prove that for each positive integer $n$, there exists a normal subgroup $H$ of $G$ of index $n$.

Hint.

Use the fourth (or Lattice) isomorphism theorem.

Proof.

Let $n$ be a positive integer.
Since $G/N$ is a cyclic group, let $g$ be a generator of $G/N$.
So we have $G/N=\langle g\rangle$.
Then $\langle g^n \rangle$ is a subgroup of $G/N$ of index $n$.

By the fourth isomorphism theorem, every subgroup of $G/N$ is of the form $H/N$ for some subgroup $H$ of $G$ containing $N$.
Thus we have $\langle g^n \rangle=H/N$ for some subgroup $H$ in $G$ containing $N$.

Since $G/N$ is cyclic, it is in particular abelian.
Thus $H/N$ is a normal subgroup of $G/N$.

The fourth isomorphism theorem also implies that $H$ is a normal subgroup of $G$, and we have
\begin{align*}
[G:H]=[G/N : H/N]=n.
\end{align*}
Hence $H$ is a normal subgroup of $G$ of index $n$.

The post If the Quotient is an Infinite Cyclic Group, then Exists a Normal Subgroup of Index $n$ first appeared on Problems in Mathematics.

Let $T:\R^3 \to \R^3$ be the linear transformation defined by the formula
\[T\left(\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \,\right)=\begin{bmatrix}
x_1+3x_2-2x_3 \\
2x_1+3x_2 \\
x_2-x_3
\end{bmatrix}.\]

Determine whether $T$ is an isomorphism and if so find the formula for the inverse linear transformation $T^{-1}$.

Solution.

Let $B=\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\}$ be the standard basis of $\R^3$, where
\[\mathbf{e}_1=\begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}, \mathbf{e}_2=\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}, \mathbf{e}_3=\begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}.\]

We determine the matrix representation $[T]_B$ of $T$ with respect to the basis $B$.
Since we have
\begin{align*}
T(\mathbf{e}_1)=\begin{bmatrix}
1 \\
2 \\
0
\end{bmatrix}, T(\mathbf{e}_2)=\begin{bmatrix}
3 \\
3 \\
1
\end{bmatrix}, T(\mathbf{e}_3=\begin{bmatrix}
-2 \\
0 \\
-1
\end{bmatrix},
\end{align*}
we have
\[[T]_B=\begin{bmatrix}
T(\mathbf{e}_1) & T(\mathbf{e}_2) & T(\mathbf{e}_3)
\end{bmatrix}=\begin{bmatrix}
1 & 3 & -2 \\
2 &3 &0 \\
0 & 1 & -1
\end{bmatrix}\\.\]

This matrix is invertible and the inverse matrix is given by
\[[T]_B^{-1}=\begin{bmatrix}
3 & -1 & -6 \\
-2 &1 &4 \\
-2 & 1 & 3
\end{bmatrix}.\] (See the post Find the Inverse Matrices if Matrices are Invertible by Elementary Row Operations for details of how to find the inverse matrix of this matrix.)

This implies that the matrix $T$ is an isomorphism.

Observe that we have $[T]_B^{-1}=[T^{-1}]_B$.
Thus, we obtain
\begin{align*}
T^{-1}\left(\, \begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix} \,\right)&=[T^{-1}]_B\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}
\\[6pt] &=\begin{bmatrix}
3 & -1 & -6 \\
-2 &1 &4 \\
-2 & 1 & 3
\end{bmatrix}\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix}\\[6pt] &=\begin{bmatrix}
3x_1-x_2-6x_3 \\
-2x_1+x_2+4x_3 \\
-2x_1+x_2+3x_3
\end{bmatrix}.
\end{align*}

In summary, the formula for the inverse linear transformation $T^{-1}$ is given by

The post Find the Inverse Linear Transformation if the Linear Transformation is an Isomorphism first appeared on Problems in Mathematics.

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.

Prove that every finite group having more than two elements has a nontrivial automorphism.

(Michigan State University, Abstract Algebra Qualifying Exam)
 

Proof.

Let $G$ be a finite group and $|G|> 2$.

Case When $G$ is a Non-Abelian Group

Let us first consider the case when $G$ is a non-abelian group.
Then there exist elements $g, h\in G$ such that $gh\neq hg$.

Consider the map $\phi: G \to G$ defined by sending $x\in G$ to $gxg^{-1}$.
Then it is straightforward to check that $\phi$ is a group homomorphism and its inverse is given by the conjugation by $g^{-1}$.
Hence $\phi$ is an automorphism.

If $\phi=1$, then we have $h=\phi(h)=ghg^{-1}$, and this implies that $gh=hg$.
This contradicts our choice of $g$ and $h$.
Hence $\phi$ is a non-trivial automorphism of $G$.

Case When $G$ is an Abelian Group

Next consider the case when $G$ is a finite abelian group of order greater than $2$.
Since $G$ is an abelian group the map $\psi:G\to G$ given by $x \mapsto x^{-1}$ is an isomorphism, hence an automorphism.

If $\psi$ is a trivial automorphism, then we have $x=\psi(x)=x^{-1}$.
Thus, $x^2=e$, where $e$ is the identity element of $G$.

Sub-Case When $G$ has an Element of Order $\geq 3$.

Therefore, if $G$ has at least one element of order greater than $2$, then $\psi$ is a non-trivial automorphism.

Sub-Case When Elements of $G$ has order $\leq 2$.

It remains to consider the case when $G$ is a finite abelian group such that $x^2=e$ for all elements $x\in G$.
In this case, the group $G$ is isomorphic to
\[\Zmod{2}\times \Zmod{2}\times \cdots \Zmod{2}=(\Zmod{2})^n.\] Since $|G| > 2$, we have $n>1$.

Then the map $(\Zmod{2})^n\to (\Zmod{2})^n$ defined by exchanging the first two entries
\[(x_1, x_2, x_3, \dots, x_n) \mapsto (x_2, x_1, x_3, \dots, x_n)\] is an example of nontrivial automorphism of $G$.

Therefore, in any case, the group $G$ has a nontrivial automorphism.

The post Every Finite Group Having More than Two Elements Has a Nontrivial Automorphism first appeared on Problems in Mathematics.

Prove that the rings $\Z[x]$ and $\Q[x]$ are not isomoprhic.

Proof.

We give three proofs.
The first two proofs use only the properties of ring homomorphism.

The third proof resort to the units of rings.

If you are familiar with units of $\Z[x]$, then the third proof might be concise and easy to follow.

The First Proof

Assume on the contrary that the rings $\Z[x]$ and $\Q[x]$ are isomorphic.
Let
\[\phi:\Q[x] \to \Z[x]\] be an isomorphism.

The polynomial $x$ in $\Q[x]$ is mapped to the polynomial $\phi(x)\in \Z[x]$.

Note that $\frac{x}{2^n}$ is an element in $\Q[x]$ for any positive integer $n$.
Thus we have
\begin{align*}
\phi(x)&=\phi(2^n\cdot \frac{x}{2^n})\\
&=2^n\phi\left(\frac{x}{2^n}\right)
\end{align*}
since $\phi$ is a homomorphism.

As $\phi$ is injective, the polynomial $\phi(\frac{x}{2^n})\neq 0$.
Since $\phi(\frac{x}{2^n})$ is a nonzero polynomial with integer coefficients, the absolute values of the nonzero coefficients of $2^n\phi(\frac{x}{2^n})$ is at least $2^n$.

However, since this is true for any positive integer, the coefficients of the polynomial $\phi(x)=2^n\phi(\frac{x}{2^n})$ is arbitrarily large, which is impossible.
Thus, there is no isomorphism between $\Q[x]$ and $\Z[x]$.

The Second Proof

Seeking a contradiction, assume that we have an isomorphism
\[\phi:\Q[x] \to \Z[x].\]

Since $\phi$ is a ring homomorphism, we have $\phi(1)=1$.
Then we have
\begin{align*}
1&=\phi(1)=\phi \left(2\cdot \frac{1}{2}\right)\\
&=2\phi\left( \frac{1}{2} \right)
\end{align*}
since $\phi$ is a homomorphism.

Since $\phi\left( \frac{1}{2} \right)\in \Z[x]$, we write
\[\phi\left( \frac{1}{2} \right)=a_nx^n+a_{n-1}x^{n-1}+\cdots a_1x+a_0,\] for some integers $a_0, \dots, a_n$.

Since $2\phi\left( \frac{1}{2} \right)=1$, it follows that
\[2a_n=0, \dots, 2a_1=0, 2a_0=1.\] Since $a_0$ is an integer, this is a contradiction.
Thus, such an isomorphism does not exists.
Hence $\Q[x]$ and $\Z[x]$ are not isomorphic.

The Third Proof

Note that in general the units of the polynomial ring $R[x]$ over an integral domain $R$ is the units $R^{\times}$ of $R$.

Since $\Z$ and $\Q$ are both integral domain, the units are
\[\Z[x]^{\times}=\Z^{\times}=\{\pm 1\} \text{ and } \Q[x]^{\times}=\Q^{\times}=\Q\setminus \{0\}.\] Since every ring isomorphism maps units to units, if two rings are isomorphic then the number of units must be the same.

As seen above, $\Z[x]$ contains only two units although $\Q[x]$ contains infinitely many units.
Thus, they cannot be isomorphic.

The post The Polynomial Rings $\Z[x]$ and $\Q[x]$ are Not Isomorphic first appeared on Problems in Mathematics.

Let $G$ and $H$ be groups and let $\phi: G \to H$ be a group homomorphism.
Suppose that $f:G\to H$ is bijective.
Then there exists a map $\psi:H\to G$ such that
\[\psi \circ \phi=\id_G \text{ and } \phi \circ \psi=\id_H.\] Then prove that $\psi:H \to G$ is also a group homomorphism.

Proof.

Let $a, b$ be arbitrary elements of the group $H$.
To prove $\psi: H \to G$ is a group homomorphism, we need
\[\psi(ab)=\psi(a)\psi(b).\]

We compute
\begin{align*}
&\phi\left(\, \psi(a)\psi(b) \,\right)\\
&=\phi\left(\, \psi(a) \,\right) \phi\left(\, \psi(b) \,\right) && \text{since $\phi$ is a group homomorphism}\\
&=ab && \text{since $\phi\circ \psi=\id_H$}\\
&=\phi \left(\, \psi (ab) \,\right) && \text{since $\phi\circ \psi=\id_H$}.\\
\end{align*}

Since $\phi$ is injective, it yields that
\[\psi(ab)=\psi(a)\psi(b),\] and thus $\psi:H\to G$ is a group homomorphism.

What’s an Isomorphism?

A bijective group homomorphism $\phi:G \to H$ is called isomorphism.

The above problem guarantees that the inverse map of an isomorphism is again a homomorphism, and hence isomorphism.

The post Inverse Map of a Bijective Homomorphism is a Group Homomorphism 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.