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On the centralizer

Suppose that A \in \mathcal{M}_n(\mathbb{C}) has this property that if \lambda is an eigenvalue of A then -\lambda is not an eigenvalue of A. Show that AX=XA if and only if A^2X=XA^2 for any X \in \mathcal{M}_n(\mathbb{C}). In other words the centralizer of A equals the centralizer of A^2.


It is clear that AX=XA implies A^2X=XA^2 for any X \in \mathcal{M}_n(\mathbb{C}). Now suppose that A^2X=XA^2 for some X \in \mathcal{M}_n(\mathbb{C}) and set Y=AX-XA. We want to prove that Y=0. We have

    \begin{align*}AY+YA &=A(AX-XA)+(AX-XA)A\\ &=A^2X-XA^2\\ &=0 \end{align*}

and so AY=-YA. It now follows that A^kY=(-1)^kYA^k for any integer k \geq 0 and thus for any \lambda \in \mathbb{C} and any integer m \geq 0 we have

    \begin{align*}(A+\lambda \mathbb{I})^mY &=\sum_{k=0}^m\binom{m}{k}\lambda^{m-k}A^kY\\ &=\sum_{k=0}^m\binom{m}{k}(-1)^k\lambda^{m-k}YA^k \\ &=(-1)^mY\sum_{k=0}^m\binom{m}{k}(-\lambda)^{m-k}A^k \\ &=(-1)^mY(A-\lambda \mathbb{I}_{n})^m \quad (*) \end{align*}

where \mathbb{I} is the identity matrix. Now let v be a generalized eigenvector corresponding to an eigenvalue \lambda of A. Then (A-\lambda \mathbb{I}_{n})^mv=0 for some integer m and thus, by (*) we have (A+\lambda \mathbb{I}_n)^mYv=0. Therefore, since we are assuming that -\lambda is not an eigenvalue of A, we must have Yv=0. So, since every element of \mathbb{C}^n is a linear combination of some generalized eigenvectors of A, we get Yu=0 for all u \in \mathbb{C}^n, i.e. Y=0 and hence AX=XA.

The exercise can also be found here.

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Isomorphic groups

Let n>2 . Define the group

    \[\mathcal{Q}_{2^n} = \langle x, y \mid x^2=y^{2^{n-2}} , y^{2^{n-1}} = 1, x^{-1} yx =y^{-1} \rangle\]

Prove that \displaystyle \mathcal{Q}_{2^n} / \mathcal{Z} \left ( \mathcal{Q}_{2^n} \right ) \simeq \mathcal{D}_{2^{n-1}} where \mathcal{D} is the dihedral group.


Using  x^{-1}yx = y^{-1} or equivalently yx = xy^{-1} we can write each element of  \mathcal{Q}_{2^n} in the form x^ry^s where r,s \in \mathbb{N} \cup \{0\}. Using x^2 = y^{2^{n-2}} we may assume that r\in \{0,1\}. Using y^{2^{n-1}} = 1 we may also assume that s\in \{0,1,\ldots,2^{n-1}-1\}. It is easy to prove inductively that y^tx = xy^{-t}.

Let \mathcal{Z} = \mathcal{Z}(\mathcal{Q}_{2^n}). We prove that \mathcal{Z}= \{1,y^{2^{n-2}}\}. Obviously 1 \in \mathcal{Z}. Furthermore,  y^{2^{n-2}} \in \mathcal{Z} since

    \[y^{2^{n-2}}\left(x^ry^s\right) = xy^{-2^{n-2}}x^{r-1}y^s = \cdots = x^ry^{2^{n-2}}y^s =\left(x^ry^s\right)y^{2^{n-2}}\]

If y^k \in \mathcal{Z} (such that 0 \leq k < 2^{n-1}) then xy^k = y^kx = xy^{-k} hence y^{2k} = 1 and therefore k = 0 or k = 2^{n-2}. If xy^k \in \mathcal{Z} then xy^{k+1} = yxy^{k} = xy^{k-1} hence y^2 = 1 which is a contradiction since n \geq 2.


    \[\mathcal{Q}_{2^n}/\mathcal{Z} = \langle x,y| x^2 = y^{2^{n-2}}=1, yx = xy^{-1} \rangle\]

which is precisely the dihedral group with 2^{n-1} elements.

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Inequality on groups

Let \mathcal{G} be a finite group and suppose that \mathcal{H} , \mathcal{K} are two subgroups of \mathcal{G} such that \mathcal{H} \neq \mathcal{G} and \mathcal{K} \neq \mathcal{G}. Show that

    \[\left|\mathcal{H} \cup \mathcal{K} \right| \leq \frac{3}{4} \left| \mathcal{G} \right|\]


Recall that \displaystyle |\mathcal{H}\mathcal{K}|=\frac{|\mathcal{H}||\mathcal{K}|}{|\mathcal{H} \cap \mathcal{K}|} and thus \displaystyle \frac{|\mathcal{H}||\mathcal{K}|}{|\mathcal{H} \cap \mathcal{K}|} \leq |\mathcal{G}|. Hence \displaystyle |\mathcal{H} \cap \mathcal{K}| \ge \frac{|\mathcal{H}| |\mathcal{K}|}{|\mathcal{G}|} and so

(1)   \begin{align*} |\mathcal{H} \cup \mathcal{K}|&=|\mathcal{H}|+|\mathcal{K}|-|\mathcal{H} \cap \mathcal{K}| \\ &\leq |\mathcal{H}|+|\mathcal{K}|-\frac{|\mathcal{H}| |\mathcal{K}|}{|\mathcal{G}|} \\ & =(a+b-ab)|\mathcal{G}|  \end{align*}

where \displaystyle a=\frac{|\mathcal{H}|}{|\mathcal{G}|} and \displaystyle b=\frac{|\mathcal{\mathcal{K}}|}{|\mathcal{G}|}.

Now, since \mathcal{H} \neq \mathcal{G} and \mathcal{K} \neq \mathcal{G} we have [\mathcal{G}: \mathcal{H}] \geq 2 and [\mathcal{G}:\mathcal{K}] \geq 2 that is a \leq \frac{1}{2} and b \leq \frac{1}{2}. So if we let a'=1-2a and b'=1-2b then a', b' \geq 0 and thus

    \[a+b-ab=\frac{3}{4}-\frac{a'+b'+a'b'}{4} \leq \frac{3}{4}\]

due to (1).

The exercise along its solution have been migrated from here .

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Irreducible factors of a polynomial

Let n \geq 1 and let

    \[p_n(x)=x^{2^n}+x^{2^{n-1}}+1 \in \mathbb{Z}[x]\]

Find all irreducible factors of p_n(x).


Setting q_n(x)=x^{2^n}-x^{2^{n-1}}+1 we note that

    \[p_n(x)=p_{n-1}(x)q_{n-1}(x) \quad , \quad n \geq 2\]


(1)   \begin{equation*} p_n(x)=p_1(x)q_1(x)q_2(x) \cdots q_{n-1}(x)  \end{equation*}

It’s clear that p_1(x)=x^2+x+1 is irreducible over \mathbb{Z}. Now, for n \geq 1 let \Phi_n(x) be the n-th cyclotomic polynomial. Using well-known properties of \Phi_n, we have

    \[\Phi_{3 \cdot 2^n}(x)=\frac{\Phi_{2^n}(x^3)}{\Phi_{2^n}(x)}=\frac{x^{3 \cdot 2^{n-1}}+1}{x^{2^{n-1}}+1}=x^{2^n}-x^{2^{n-1}}+1=q_n(x)\]

Thus q_n is irreducible over \mathbb{Z} because cyclotomic polynomials are irreducible over \mathbb{Z}. Hence, by ( 1 ) p_n has exactly n irreducible factors and they are p_1, q_1, q_2, \dots , q_{n-1}.

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Galois theory … of the Euler’s totient function

Let n>2 and let \omega \in \mathbb{C} be an n-th primitive root of unity. Prove that

    \[[\mathbb{Q}(\omega + \omega^{-1}) :\mathbb{Q}]=\frac{\varphi (n)}{2}\]

where \phi denotes the Euler’s totient function.


We have

    \[\phi(n)=[\mathbb{Q}(\omega):\mathbb{Q}]=[\mathbb{Q}(\omega):\mathbb{Q}(\omega +\omega^{-1})][\mathbb{Q}(\omega +\omega^{-1}):\mathbb{Q}]\]

where [\mathbb{Q}(\omega):\mathbb{Q}(\omega +\omega^{-1})]=2 since t^2-(\omega+\omega^{-1})t+1 is the minimal polynomial of \omega over \mathbb{Q}(\omega+\omega^{-1}).

The result follows.

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