No Finite Abelian Group is Divisible

Problem 240

A nontrivial abelian group $A$ is called divisible if for each element $a\in A$ and each nonzero integer $k$, there is an element $x \in A$ such that $x^k=a$.
(Here the group operation of $A$ is written multiplicatively. In additive notation, the equation is written as $kx=a$.) That is, $A$ is divisible if each element has a $k$-th root in $A$.

(a) Prove that the additive group of rational numbers $\Q$ is divisible.

(b) Prove that no finite abelian group is divisible.

 
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Orthogonality of Eigenvectors of a Symmetric Matrix Corresponding to Distinct Eigenvalues

Problem 235

Suppose that a real symmetric matrix $A$ has two distinct eigenvalues $\alpha$ and $\beta$.
Show that any eigenvector corresponding to $\alpha$ is orthogonal to any eigenvector corresponding to $\beta$.

(Nagoya University, Linear Algebra Final Exam Problem)
 
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Explicit Field Isomorphism of Finite Fields

Problem 233

(a) Let $f_1(x)$ and $f_2(x)$ be irreducible polynomials over a finite field $\F_p$, where $p$ is a prime number. Suppose that $f_1(x)$ and $f_2(x)$ have the same degrees. Then show that fields $\F_p[x]/(f_1(x))$ and $\F_p[x]/(f_2(x))$ are isomorphic.

(b) Show that the polynomials $x^3-x+1$ and $x^3-x-1$ are both irreducible polynomials over the finite field $\F_3$.

(c) Exhibit an explicit isomorphism between the splitting fields of $x^3-x+1$ and $x^3-x-1$ over $\F_3$.

 
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Polynomial $x^p-x+a$ is Irreducible and Separable Over a Finite Field

Problem 229

Let $p\in \Z$ be a prime number and let $\F_p$ be the field of $p$ elements.
For any nonzero element $a\in \F_p$, prove that the polynomial
\[f(x)=x^p-x+a\] is irreducible and separable over $F_p$.

(Dummit and Foote “Abstract Algebra” Section 13.5 Exercise #5 on p.551)

 
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Group of $p$-Power Roots of 1 is Isomorphic to a Proper Quotient of Itself

Problem 221

Let $p$ be a prime number. Let
\[G=\{z\in \C \mid z^{p^n}=1\} \] be the group of $p$-power roots of $1$ in $\C$.

Show that the map $\Psi:G\to G$ mapping $z$ to $z^p$ is a surjective homomorphism.
Also deduce from this that $G$ is isomorphic to a proper quotient of $G$ itself.

 
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