Prove that every group of order $24$ has a normal subgroup of order $4$ or $8$.

Proof.

Let $G$ be a group of order $24$.
Note that $24=2^3\cdot 3$.
Let $P$ be a Sylow $2$-subgroup of $G$. Then $|P|=8$.

Consider the action of the group $G$ on the left cosets $G/P$ by left multiplication.
This induces a permutation representation homomorphism
\[\phi: G\to S_{G/P},\] where $S_{G/P}$ is a group of bijective maps (permutations) on $G/P$.

This homomorphism is defined by
\[\phi(g)(aP)=gaP\] for $g\in G$ and $aP\in G/P$.

Then by the first isomorphism theorem, we see that
\[G/\ker(\phi) \cong \im(\phi) < S_{G/P}.\] This implies that the order of $G/\ker(\phi)$ divides the order of $S_{G/P}$. Note that as $|G/P|=3$, we have $|S_{G/P}|=|S_3|=6$. Thus, we must have $4\mid |\ker{\phi}|$.

Also note that $\ker(\phi) < P$. To see this let $x\in \ker(\phi)$. Then we have \[xP=\phi(x)(P)=\id(P)=P.\] Here $\id$ is the identity map from $G/P$ to itself. Hence $x\in P$. It follows that $|\ker(\phi)|$ divides $|P|=8$.

Combining these restrictions, we see that $|\ker(\phi)|=4, 8$.
Being the kernel of a homomorphism, $\ker(\phi)$ is a normal subgroup of $G$.
Hence the group $G$ of order $24$ has a normal subgroup of order $4$ or $8$.

The post Every Group of Order 24 Has a Normal Subgroup of Order 4 or 8 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
\[R=\left\{\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \quad \middle | \quad a, b\in \Q \,\right\}.\] Then the usual matrix addition and multiplication make $R$ an ring.

Let
\[J=\left\{\, \begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix} \quad \middle | \quad b \in \Q \,\right\}\] be a subset of the ring $R$.

(a) Prove that the subset $J$ is an ideal of the ring $R$.

(b) Prove that the quotient ring $R/J$ is isomorphic to $\Q$.

Proof.

(a) Prove that the subset $J$ is an ideal of the ring $R$.

Let
\[\alpha=\begin{bmatrix}
0 & a\\
0& 0
\end{bmatrix} \text{ and } \beta=\begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix}\] be arbitrary elements in $J$ with $a, b\in \Q$.
Then since we have
\begin{align*}
\alpha-\beta=\begin{bmatrix}
0 & a-b\\
0& 0
\end{bmatrix}\in J,
\end{align*}
the subset $J$ is an additive group.

Now consider any elements
\[\rho=\begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \in R \text{ and } \gamma=\begin{bmatrix}
0 & c\\
0& 0
\end{bmatrix}\in J.\] Then we have
\begin{align*}
\rho \gamma&=\begin{bmatrix}
0 & ac\\
0& 0
\end{bmatrix}\in J \text{ and }\\[6pt] \gamma \rho &=\begin{bmatrix}
0 & ca\\
0& 0
\end{bmatrix}\in J.
\end{align*}
Thus, each element of $J$ multiplied by an element of $R$ is still in $J$.

Hence $J$ is an ideal of the ring $R$.

(b) Prove that the quotient ring $R/J$ is isomorphic to $\Q$.

Consider the map $\phi:R\to \Q$ defined by
\[\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \,\right)=a,\] for $\begin{bmatrix}
a & b\\
0& a
\end{bmatrix}\in R$.

We first show that the map $\phi$ is a ring homomorphism.
First of all, we have
\begin{align*}
\phi\left(\, \begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix} \,\right)=1.
\end{align*}
Thus $\phi$ maps the unity element of $R$ to the unity element of $\Q$.

Take
\[\begin{bmatrix}
a & b\\
0& a
\end{bmatrix}, \begin{bmatrix}
c & d\\
0& c
\end{bmatrix}\in R.\] Then we have
\begin{align*}
\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix}+\begin{bmatrix}
c & d\\
0& c
\end{bmatrix} \,\right)&=\phi\left(\, \begin{bmatrix}
a+c & b+d\\
0& a+c
\end{bmatrix} \,\right)=a+c\\[6pt] &=\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \,\right)+\phi\left(\, \begin{bmatrix}
c & d\\
0& c
\end{bmatrix} \,\right)
\end{align*}
and
\begin{align*}
\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix}\begin{bmatrix}
c & d\\
0& c
\end{bmatrix} \,\right)&=\phi\left(\, \begin{bmatrix}
ac & ad+bc\\
0& ac
\end{bmatrix} \,\right)=ac\\[6pt] &=\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \,\right)\phi\left(\, \begin{bmatrix}
c & d\\
0& c
\end{bmatrix} \,\right).
\end{align*}
It follows from these computations that $\phi:R\to \Q$ is a ring homomorphism.

Next, we determine the kernel of $\phi$.
We claim that $\ker(\phi)=J$.

If $\rho=\begin{bmatrix}
a & b\\
0& a
\end{bmatrix}\in \ker(\phi)$, then we have
\[0=\phi(\rho)=\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix} \,\right)=a.\]

So $\rho=\begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix}\in J$, and hence $\ker(\phi)\subset J$.

On the other hand, if $\rho=\begin{bmatrix}
0 & b\\
0& 0
\end{bmatrix}\in J$, then it follows from the definition of $\phi$ that $\phi(\rho)=0$.
Thus, $J \subset \ker(\phi)$.
Putting these two inclusions together yields $J=\ker(\phi)$.

Observe that the homomorphism $\phi$ is surjective.
In fact, for any $a\in \Q$, we take $\begin{bmatrix}
a & 0\\
0& a
\end{bmatrix}\in R$. Then we have
\[\phi\left(\, \begin{bmatrix}
a & 0\\
0& a
\end{bmatrix} \,\right)=a.\]

In summary, $\phi:R\to \Q$ is a surjective homomorphism with kernel $J$.

It follows from the isomorphism theorem that
\[R/J\cong \Q,\] as required.

Remark.

Recall that the kernel of a ring homomorphism $\phi:R\to S$ is always an ideal of $R$.

Thus, the proof of (b) shows that $J$ is an ideal of $R$. This gives an alternative proof of part (a).

The post The Quotient Ring by an Ideal of a Ring of Some Matrices is Isomorphic to $\Q$. first appeared on Problems in Mathematics.

Let $R$ be a commutative ring. Consider the polynomial ring $R[x,y]$ in two variables $x, y$.
Let $(x)$ be the principal ideal of $R[x,y]$ generated by $x$.

Prove that $R[x, y]/(x)$ is isomorphic to $R[y]$ as a ring.

Proof.

Define the map $\psi: R[x,y] \to R[y]$ by sending $f(x,y)\in R[x,y]$ to $f(0,y)$.
Namely, the map $\psi$ is the substitution $x=0$.
It is straightforward to check that $\psi$ is a ring homomorphism.

For any polynomial $g(y)\in R[y]$, let $G(x, y)=g(y)\in R[x,y]$.
Then we have $\psi(G(x,y))=G(0,y)=g(y)$.
This proves that $\psi$ is surjective.

We claim that the kernel of $\psi$ is the ideal $(x)$.

Suppose that $f(x,y) \in \ker(\psi)$.
We write
\[f(x,y)=f_0(y)+f_1(y)x+\cdots +f_n(y)x^n,\] where $f_i\in R[y]$ for $i=1, \dots, n$.

Since $f(x,y)\in \ker(\psi)$, it yields that
\[0=\psi(f(x,y))=f(0,y)=f_0(y).\] Hence
\begin{align*}
f(x,y)&=f_1(y)x+\cdots +f_n(y)x^n\\
&=x\left(f_1(y)+\cdots +f_n(y)x^{n-1}\right) \in (x).
\end{align*}
Thus, $\ker(\psi) \subset (x)$.

On the other hand, suppose $f(x,y)\in (x)$.
Then there exists $g(x,y) \in R[x,y]$ such that
\[f(x,y)=xg(x,y).\] It follows that
\begin{align*}
\psi\left(\, f(x,y) \,\right)&=\psi\left(\, xg(x,y) \,\right)=0g(0,y)=0.
\end{align*}
It implies that $f(x,y) \in \ker(\psi)$, hence $\ker(\psi) \subset (x)$.

Putting two inclusions together gives $(x)=\ker(\psi)$.

In summary, the map $\psi:R[x,y] \to R[y]$ is a surjective ring homomorphism with kernel $(x)$.

Hence by the isomorphism theorem, we obtain the isomorphism
\[R[x,y]/(x)\cong R[y].\] Click here if solved 21

The post Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$ first appeared on Problems in Mathematics.

Let $R$ be a commutative ring with $1$ and let $M$ be an $R$-module.
Prove that the $R$-module $M$ is irreducible if and only if $M$ is isomorphic to $R/I$, where $I$ is a maximal ideal of $R$, as an $R$-module.

Definition (Irreducible module).

An $R$-module $M$ is called irreducible if $M$ is not the zero module and $0$ and $M$ are the only submodules of $M$.

An irreducible module
is also called a simple module.

Proof.

$(\implies)$ Suppose that $M$ is an irreducible $R$-module.
Then by definition of an irreducible module, $M$ is not the zero module.
Take any nonzero element $m\in M$, and consider the cyclic submodule $(m)=Rm$ generated by $m$.
Since $M$ is irreducible, we must have $M=Rm$.

Now we define a map $f:R\to M$ by sending $r\in R$ to $f(r)=rm$.
Then the map $f$ is an $R$-module homomorphism regarding $R$ is an $R$-module.

In fact, we have
\begin{align*}
&f(r+s)=(r+s)m=rm+sm=f(r)+f(s)\\
&f(rs)=(rs)m=r(sm)=rf(s)
\end{align*}
for any $r, s\in R$.
Since $M=Rm$, the homomorphism $f$ is surjective.
Thus, by the first isomorphism theorem, we obtain
\[R/I\cong M,\] where $I=\ker(f)$.

It remains to show that $I$ is a maximal ideal of $R$.
Suppose that $J$ is an ideal such that $I\subset J \subset R$.
Then by the third isomorphism theorem for rings, we know that $J/I$ is an ideal of the ring $R/I\cong M$, hence $J/I$ is a submodule.

Since $M$ is irreducible, we must have either $J/I=0$ or $J/I=M$.
This implies that $J=I$ or $J=M$.
Hence $I$ is a maximal ideal.

$(\impliedby)$ Suppose now that $M\cong R/I$ for some maximal ideal $I$ of $R$.
Let $N$ be any submodule of $R/I$. (We identified $M$ and $R/I$ by the above isomorphism.)
Then $N$ is an ideal of $R/I$ since $N$ is an abelian group and closed under the action of $R$, hence that of $R/I$.

Since $R$ is a commutative ring and $I$ is a maximal ideal of $R$, we know that $R/I$ is a field.
Thus, only the ideals of $R/I$ are $0$ or $R/I$.

Hence we have $N=0$ or $N=R/I=M$.
This proves that $M$ is irreducible.

Related Question.

A similar technique in the proof above can be used to solve the following problem.

See the post “A module is irreducible if and only if it is a cyclic module with any nonzero element as generator” for a proof.

The post A Module $M$ is Irreducible if and only if $M$ is isomorphic to $R/I$ for a Maximal Ideal $I$. first appeared on Problems in Mathematics.

Let $\calF[0, 2\pi]$ be the vector space of all real valued functions defined on the interval $[0, 2\pi]$.
Define the map $f:\R^2 \to \calF[0, 2\pi]$ by
\[\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix} \,\right) \,\right)(x):=\alpha \cos x + \beta \sin x.\] We put
\[V:=\im f=\{\alpha \cos x + \beta \sin x \in \calF[0, 2\pi] \mid \alpha, \beta \in \R\}.\]

(a) Prove that the map $f$ is a linear transformation.

(b) Prove that the set $\{\cos x, \sin x\}$ is a basis of the vector space $V$.

(c) Prove that the kernel is trivial, that is, $\ker f=\{\mathbf{0}\}$.
(This yields an isomorphism of $\R^2$ and $V$.)

(d) Define a map $g:V \to V$ by
\[g(\alpha \cos x + \beta \sin x):=\frac{d}{dx}(\alpha \cos x+ \beta \sin x)=\beta \cos x -\alpha \sin x.\] Prove that the map $g$ is a linear transformation.

(e) Find the matrix representation of the linear transformation $g$ with respect to the basis $\{\cos x, \sin x\}$.

(Kyoto University, Linear Algebra exam problem)

Terminology (image=range, kernel=null space)

We remark that the image is also called the range, and the kernel is also called the null space.

Proof.

(a) Prove that the map $f$ is a linear transformation.

To prove that $f:\R^2 \to \calF[0, 2\pi]$ is a linear transformation, we check the following two properties:

1. \[f\left(\, \begin{bmatrix}
   \alpha \\
   \beta
   \end{bmatrix}+\begin{bmatrix}
   \alpha’ \\
   \beta’
   \end{bmatrix} \,\right)=f\left(\, \begin{bmatrix}
   \alpha \\
   \beta
   \end{bmatrix}\,\right)+f\left(\, \begin{bmatrix}
   \alpha’ \\
   \beta’
   \end{bmatrix} \,\right)\] for any vectors $\begin{bmatrix}
   \alpha\\
   \beta
   \end{bmatrix}, \begin{bmatrix}
   \alpha’ \\
   \beta’
   \end{bmatrix} \in \R^2$.
2. \[f\left(\, r\begin{bmatrix}
   \alpha \\
   \beta
   \end{bmatrix} \,\right)=rf\left(\, \begin{bmatrix}
   \alpha \\
   \beta
   \end{bmatrix}\,\right)\] for any vector $\begin{bmatrix}
   \alpha\\
   \beta
   \end{bmatrix} \in \R^2$ and any scalar $r\in \R$.

For any vectors $\begin{bmatrix}
\alpha\\
\beta
\end{bmatrix}, \begin{bmatrix}
\alpha’ \\
\beta’
\end{bmatrix} \in \R^2$, we have
\begin{align*}
\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}+\begin{bmatrix}
\alpha’ \\
\beta’
\end{bmatrix} \,\right) \,\right)(x)&=
\left(\, f\left(\, \begin{bmatrix}
\alpha+\alpha’ \\
\beta+\beta’
\end{bmatrix} \,\right) \,\right)(x)\\[6pt] &=(\alpha+\alpha’) \cos x + (\beta+\beta’) \sin x\\
&=(\alpha \cos x + \beta \sin x)+(\alpha’\cos x + \beta’ \sin x)\\
&=\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix} \,\right) \,\right)(x)+\left(\, f\left(\, \begin{bmatrix}
\alpha’ \\
\beta’
\end{bmatrix} \,\right) \,\right)(x),
\end{align*}
and hence the first property is proved.

For the second one, let $\begin{bmatrix}
\alpha\\
\beta
\end{bmatrix} \in \R^2$ and $r\in \R$.
We have
\begin{align*}
\left(\, f\left(\, r\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix} \,\right) \,\right)(x)&=
\left(\, f\left(\, \begin{bmatrix}
r\alpha \\
r\beta
\end{bmatrix} \,\right) \,\right)(x)\\
&=(r\alpha) \cos x + (r\beta) \sin x\\
&=r(\alpha \cos x + \beta \sin x)\\
&=r\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}\,\right) \,\right)(x).
\end{align*}
This proves the property 2, and hence the map $f$ is a linear transformation.

(b) Prove that the set $\{\cos x, \sin x\}$ is a basis of the vector space $V$.

It is clear that the set $\{\cos x, \sin x\}$ spans the vector space $V$ by definition. Thus, it suffices to prove that $\{\cos x, \sin x\}$ is a linearly independent set.
Suppose that we have
\[\alpha \cos x + \beta \sin x=0\] for some $\alpha, \beta \in \R$.
This is an equality as vectors in $V$, that is, this equality holds for all $x\in [0, 2\pi]$.

In particular, we substitute $x=0$ and obtain $\alpha=0$.
Also, we substitute $x=\pi/2$ and obtain $\beta=0$.
It follows that $\{\cos x, \sin x\}$ is linearly independent, and hence it is a basis of $V$.

(c) Prove that the kernel $\ker f=\{\mathbf{0}\}$.

If $\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}\in \ker f$, then we have
\[\left(\, f\left(\, \begin{bmatrix}
\alpha \\
\beta
\end{bmatrix} \,\right) \,\right)(x)=\alpha \cos x + \beta \sin x=0\] for all $x\in [0, 2\pi]$.
As in part (b) this implies that $\alpha=\beta=0$.
Hence we have
\[\begin{bmatrix}
\alpha \\
\beta
\end{bmatrix}=\begin{bmatrix}
0 \\
0
\end{bmatrix}=\mathbf{0},\] and thus we obtain $\ker f=\{\mathbf{0}\}$.

By the isomorphism theorem, we obtain
\[\R^2=\R^2/\{0\}=\R^2/\ker f \cong \im f =V.\]

(d) Prove that the map $g$ is a linear transformation.

Let $\alpha \cos x + \beta \sin x, \alpha’ \cos x + \beta’ \sin x$ be arbitrary vectors in $V$. Then we have
\begin{align*}
&g\left(\, (\alpha \cos x + \beta \sin x)+(\alpha’ \cos x + \beta’ \sin x) \,\right)\\
&=g\left(\, (\alpha+\alpha’) \cos x + (\beta+\beta’) \sin x)\,\right)\\
&=(\beta+\beta’)\cos x-(\alpha+\alpha’) \sin x\\
&=(\beta \cos x -\alpha \sin x)+(\beta’ \cos x -\alpha’ \sin x)\\
&=g(\alpha \cos x + \beta \sin x)+g(\alpha’ \cos x + \beta’ \sin x).
\end{align*}

We also have for any vector $\alpha \cos x + \beta \sin x \in V$ and any scalar $r\in \R$
\begin{align*}
&g\left(\, r(\alpha \cos x + \beta \sin x) \,\right)\\
&=g\left(\, (r\alpha) \cos x + (r\beta) \sin x) \,\right)\\
&=(r\beta)\cos x – (r\alpha)\sin x\\
&=r(\beta\cos x – \alpha\sin x)\\
&=rg(\alpha \cos x + \beta \sin x).
\end{align*}
Therefore $g$ is a linear transformation.

(e) Find the matrix representation of the linear transformation $g$ with respect to the basis $\{\cos x, \sin x\}$.

We have
\begin{align*}
g(\cos x)=-\sin x \text{ and } g(\sin x)=\cos x.
\end{align*}
Thus, the coordinate vectors with respect to the basis $B:=\{\cos x, \sin x\}$ are
\[[g(\cos x)]_B=\begin{bmatrix}
0 \\
-1
\end{bmatrix} \text{ and } [g(\sin x)]_B=\begin{bmatrix}
1 \\
0
\end{bmatrix}.\] It follows that the matrix representation of $g$ is
\[ [\,[g(\cos x)]_B, [g(\sin x)]_B\,]=\begin{bmatrix}
0 & 1\\
-1& 0
\end{bmatrix}.\]

Alternatively, we can write
\[g[\cos x, \sin x]=[\cos x, \sin x] \begin{bmatrix}
0 & 1\\
-1& 0
\end{bmatrix}.\] Click here if solved 20

The post Subspace Spanned By Cosine and Sine Functions first appeared on Problems in Mathematics.

Let $A$ be an $n\times n$ idempotent matrix, that is, $A^2=A$. Then prove that $A$ is diagonalizable.

 

We give three proofs of this problem. The first one proves that $\R^n$ is a direct sum of eigenspaces of $A$, hence $A$ is diagonalizable.

The second proof proves the direct sum expression as in proof 1 but we use a linear transformation.

The third proof discusses the minimal polynomial of $A$.

Range=Image, Null space=Kernel

In the following proofs, we use the terminologies range and null space of a linear transformation. These are also called image and kernel of a linear transformation, respectively.

Proof 1.

Recall that only possible eigenvalues of an idempotent matrix are $0$ or $1$.
(For a proof, see the post “Idempotent matrix and its eigenvalues“.)

Let
\[E_0=\{\mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{0}\} \text{ and } E_{1}\{\mathbf{x}\in \R^n \mid A\mathbf{x}=\mathbf{x}\}\] be subspaces of $\R^n$.
(Thus, if $0$ and $1$ are eigenvalues, then $E_0$ and $E_1$ are eigenspaces.)

Let $r$ be the rank of $A$. Then by the rank-nullity theorem, the nullity of $A$
\[\dim(E_0)=n-r.\]

The rank of $A$ is the dimension of the range
\[\calR(A)=\{\mathbf{y} \in \R^n \mid \mathbf{y}=A\mathbf{x} \text{ for some } \mathbf{x}\in \R^n\}.\]

Let $\mathbf{y}_1, \dots, \mathbf{y}_r$ be basis vectors of $\calR(A)$.
Then there exists $\mathbf{x}_i\in \R^n$ such that
\[\mathbf{y}_i=A\mathbf{x}_i,\] for $i=1, \dots, r$.

Then we have
\begin{align*}
A\mathbf{y}_i&=A^2\mathbf{x}_i\\
&=A\mathbf{x}_i && \text{since $A$ is idempotent}\\
&=\mathbf{y}_i.
\end{align*}

It follows that $y_i\in E_1$.
Since $y_i, i=1,\dots, r$ form a basis of $\calR(A)$, they are linearly independent and thus we have
\[r\leq \dim(E_1).\]

We have
\begin{align*}
&n=\dim(\R^n)\\
&\geq \dim(E_0)+\dim(E_1) && \text{since } E_0\cap E_1=\{\mathbf{0}\}\\
&\geq (n-r)+r=n.
\end{align*}

So in fact all inequalities are equalities, and hence
\[\dim(\R^n)=\dim(E_0)+\dim(E_1).\]

This implies
\[\R^n=E_0 \oplus E_1.\] Thus $\R^n$ is a direct sum of eigenspaces of $A$, and hence $A$ is diagonalizable.

Proof 2.

Let $E_0$ and $E_1$ be as in proof 1.
Consider the linear transformation $T:\R^n \to \R^n$ represented by the idempotent matrix $A$, that is, $T(\mathbf{x})=A\mathbf{x}$.

Then the null space $\calN(T)$ of the linear transformation $T$ is $E_0$ by definition.

We claim that the range $\calR(T)$ is $E_1$.
If $\mathbf{x}\in \calR(T)$, then we have $\mathbf{y}\in \R^n$ such that $\mathbf{x}=T(\mathbf{y})=A\mathbf{y}$.

Then we have
\begin{align*}
\mathbf{x}&=A\mathbf{y}=A^2\mathbf{y} =A(A\mathbf{y})
=A\mathbf{x}.
\end{align*}
(The second equality follows since $A$ is idempotent.)

This implies that $\mathbf{x}\in E_1$, and hence $\calR(T) \subset E_1$.

On the other hand, if $\mathbf{x}\in E_1$, then we have
\[\mathbf{x}=A\mathbf{x}=T(\mathbf{x})\in \calR(T).\] Thus, we have $E_1\subset \calR(T)$. Putting these two inclusions together gives $E_1=\calR(T)$.

By the isomorphism theorem of vector spaces, we have
\[\R^n=\calN(A)\oplus \calR(T)=E_0\oplus E_1.\] Thus, $\R^n$ is a direct sum of eigenspaces of $A$ and hence $A$ is diagonalizable.

Proof 3.

Since $A$ is idempotent we have $A^2=A$.
Thus we have $A^2-A=O$, the zero matrix, and so $A$ satisfies the polynomial $x^2-x$.

If $x^2-x=x(x-1)$ is not the minimal polynomial of $A$, then $A$ must be either the identity matrix or the zero matrix.
Since these matrices are diagonalizable (as they are already diagonal matrices), we consider the case when $x^2-x$ is the minimal polynomial of $A$.

Since the minimal polynomial has two distinct simple roots $0, 1$, the matrix $A$ is diagonalizable.

Another Proof

A slightly different proof is given in the post “Idempotent (Projective) Matrices are Diagonalizable“.

The proof there is a variation of Proof 2.

The post Idempotent Matrices are Diagonalizable first appeared on Problems in Mathematics.

Let $R$ be the ring of all continuous functions on the interval $[0, 2]$.
Let $I$ be the subset of $R$ defined by
\[I:=\{ f(x) \in R \mid f(1)=0\}.\]

Then prove that $I$ is an ideal of the ring $R$.
Moreover, show that $I$ is maximal and determine $R/I$.

Hint.

Consider the map $\phi:R\to \R$ defined by
\[\phi(f)=f(1),\] for every $f(x)\in R$.

Proof.

Let us consider the map $\phi$ from $R$ to the field of real numbers $\R$ defined by
\[\phi(f)=f(1),\] for each $f(x)\in R$. Namely, the map $\phi$ is the evaluation at $x=1$.

We claim that $\phi:R \to \R$ is a ring homomorphism. In fact we have for any $f(x), g(x)\in R$,
\begin{align*}
\phi(fg)&=(fg)(1)=f(1)g(1)=\phi(f)\phi(g)\\
\phi(f+g)&=(f+g)(1)=f(1)+g(1)=\phi(f)+\phi(g),
\end{align*}
hence $\phi$ is a ring homomorphism.

Next, consider the kernel of $\phi$. We have
\begin{align*}
\ker(\phi)&=\{ f(x)\in R\mid \phi(f)=0\}\\
&=\{f(x) \in R \mid f(1)=0\}=I.
\end{align*}
Since the kernel of a ring homomorphism is an ideal, it follows that $I=\ker(\phi)$ is an ideal of $R$.

Next, we claim that $\phi$ is surjective. To see this, let $r\in \R$ be an arbitrary real number.
Define the constant function $f(x)=r$. Then $f(x)$ is an element in $R$ as it is continuous function on $[0, 2]$.
We have
\begin{align*}
\phi(f)=f(1)=r,
\end{align*}
and this proves that $\phi$ is surjective.

Since $\phi: R\to \R$ is a surjective ring homomorphism, the first isomorphism theorem yields that
\[R/\ker(\phi) \cong \R.\] Since $\ker(\phi)=I$ as we saw above, we have
\[R/I \cong \R.\] Thus, the quotient ring $R/I$ is isomorphic to the field $\R$.
It follows from this that $I$ is a maximal ideal of $R$.

(Recall the fact that an ideal $I$ of a commutative ring $R$ is maximal if and only if $R/I$ is a field.)

Related Question.

For a proof, see the post ↴
Polynomial Ring with Integer Coefficients and the Prime Ideal $I=\{f(x) \in \Z[x] \mid f(-2)=0\}$

The post A Maximal Ideal in the Ring of Continuous Functions and a Quotient Ring first appeared on Problems in Mathematics.

Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.

Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of $G$.

Outline of the proof

Here is the outline of the proof.

1. Define a group homomorphism $\psi: G\to \Aut(N)$ by $\psi(g)(n)=gng^{-1}$ for all $g\in G$ and $n\in N$.
   We need to check:
• The map $\psi(g)$ is an automorphism of $N$ for each $g\in G$.
• The map $\psi$ is in fact a group homomorphism from $G$ to $\Aut(N)$.
2. The assumption that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime implies that $G=\ker(\psi)$.
3. This implies that $N$ is in the center of $G$.

Proof.

We define a group homomorphism $\psi: G \to \Aut(N)$ as follows.
For each $g\in G$, we first define an automorphism $\psi(g)$ of $N$.
Define $\psi(g): N \to N$ by
\[\psi(g)(n)=gng^{-1}.\]

Note that since $N$ is a normal subgroup of $G$, the output $\psi(g)(n)=gng^{-1}$ actually lies in $N$.

We prove that so defined $\psi(g)$ is a group homomorphism from $N$ to $N$ for each fixed $g\in G$.
For $n_1, n_2 \in N$, we have
\begin{align*}
\psi(g)(n_1n_2)&=g(n_1n_2)g^{-1} && \text{by definition of $\psi(g)$}\\
&=gn_1g^{-1}gn_2g^{-1} && \text{by inserting $e=g^{-1}g$}\\
&=\psi(g)(n_1) \psi(g)(n_2) && \text{by definition of $\psi(g)$}.
\end{align*}
It follows that $\psi(g)$ is a group homomorphism, and hence $\psi(g)\in \Aut(N)$.

We have defined a map $\psi:G\to \Aut(N)$. We now prove that $\psi$ is a group homomorphism.
For any $g_1, g_2$, and $n\in N$, we have
\begin{align*}
\psi(g_1 g_2)(n)&=(g_1g_2)n(g_1 g_2)^{-1}\\
&=g_1 g_2 n g_2^{-1} g_1^{-1}\\
&=g_1 \psi(g_2)(n) g_1^{-1}\\
&=\psi(g_1)\psi(g_2)(n).
\end{align*}

Thus, $\psi: G\to \Aut(N)$ is a group homomorphism.
By the first isomorphism theorem, we have
\[G/\ker(\psi)\cong \im(\psi)< \Aut(N). \tag{*}\] Note that if $g\in N$, then $\psi(g)(n)=gng^{-1}=n$ since $N$ is abelian. It yields that the subgroup $N$ is in the kernel $\ker(\psi)$.

Then by the third isomorphism theorem, we have
\begin{align*}
G/\ker(\psi) \cong (G/N)/(\ker(\psi)/N). \tag{**}
\end{align*}

It follows from (*) and (**) that the order of $G/\ker(\psi)$ divides both the order of $\Aut(N)$ and the order of $G/N$. Since the orders of the latter two groups are relatively prime by assumption, the order of $G/\ker(\psi)$ must be $1$. Thus the quotient group is trivial and we have
\[G=\ker(\psi).\]

This means that for any $g\in G$, the automorphism $\psi(g)$ is the identity automorphism of $N$.
Thus, for any $g\in G$ and $n\in N$, we have $\psi(g)(n)=n$, and thus $gng^{-1}=n$.
As a result, the subgroup $N$ is contained in the center of $G$.

The post Abelian Normal subgroup, Quotient Group, and Automorphism Group first appeared on Problems in Mathematics.

Let $H$ and $K$ be normal subgroups of a group $G$.
Suppose that $H < K$ and the quotient group $G/H$ is abelian.
Then prove that $G/K$ is also an abelian group.

Solution.

We will give two proofs.

Hint (The third isomorphism theorem)

Recall the third isomorphism theorem of groups:
Let $G$ be a group and let $H, K$ be normal subgroups of $G$ with $H < K$.
Then we have $G/K$ is a normal subgroup of $G/H$ and we have an isomorphism
\[G/K \cong (G/H)/(G/K).\]

Proof 1 (Using third isomorphism theorem)

Since $H, K$ are normal subgroups of $G$ and $H < K$, the third isomorphism theorem yields that
\[G/K \cong (G/H)/(G/K).\]

Since the group $G/H$ is abelian by assumption, and in general a quotient group of an abelian group is abelian, it follows $(G/H)/(G/K)$ is an abelian group.

Hence by the above isomorphism, the group $G/K$ is also an abelian group.

Proof 2 (Using the commutator subgroup)

Here is another proof using the commutator subgroup $[G, G]$ of $G$.
Recall that for a subgroup $N$ of $G$, the following two conditions are equivalent.

1. The subgroup $N$ is normal and the $G/N$ is an abelian.
2. The commutator subgroup $[G, G]$ is a subgroup of $N$.

For the proof of this fact, see the post “Commutator subgroup and abelian quotient group“.

Now we prove the problem using this fact.
Since $H$ is normal and the quotient $G/H$ is an abelian group, the commutator subgroup $[G, G]$ is a subgroup of $H$ by the fact (1 $\implies$ 2).

Then we have
\begin{align*}
[G, G] < H < K.
\end{align*}
Hence $[G, G]$ is a subgroup of $K$, hence $G/K$ is an abelian group by the fact again (2 $\implies$ 1).

The post If Quotient $G/H$ is Abelian Group and $H < K \triangleleft　G$, then $G/K$ is Abelian first appeared on Problems in Mathematics.