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The polynomial $2x^5 - 5x^4 + 5$ is irreducible by Eisenstein's criteria using divisor $5$ given that $5 \mid a_i : i \neq 5$, $5 \nmid a_5$, and $5^2 \nmid a_0$.
$a(x)$ thus only crosses the x-axis once and has one real root $r_1$, and four complex roots $r_2, r_3, r_4, r_5$ with $r_2, r_3$ and $r_4, r_5$ being complex conjugates of each other.
The permutation group of $r_1, r_2, r_3, r_4, r_5$ which is $\mathbf{G}$ is a subgroup of $S_5$.
Since $a(x)$ is irreducible in $\mathbb{Q}$, so $[\mathbb{Q}(r_1): \mathbb{Q}] = 5$. So $5$ is a factor of $[K: \mathbb{Q}]$. Thus $\mathbf{G}$ contains an element of order 5.
The automorphism $(r_1 r_2 r_3 r_4)$ has order 4, $(r_1 r_2 r_3)$ has order 3, $(r_1 r_2 r_3)(r_4 r_5)$ order 6, which only leaves the cycle of length 5 $(r_1 r_2 r_3 r_4 r_5)$ which has order 5. Thus $\mathbf{G}$ contains an automorphism which is a cycle permutation of the roots $r_1, r_2, r_3, r_4, r_5$ of length 5.
From 8H5, we saw that a transposition $(12)$ and a cycle $(12345)$ will generate $S_5$. The proof follows: $(12345)(12)(12345)^{-1} = (23), (12)(23)(12) = (13)$. Repeating the process we get $(12345)(13)(12345)^{-1} = (24), (12)(24)(12) = (14), (12345)(14)(12345)^{-1} = (25), (12)(25)(12) = (15)$. Finally the set $T_1 = { (12), (13), \dots, (15) }$ generates $S_5$.
Thus $\mathbf{G} = S_5$ and since $S_5$ is not solvable, there is no radical solution for $a(x)$.
The graph has a maximum at $x = 0$, and a minimum at $x = \frac{2}{5} \sqrt[3]{5}^2$. $a(x)$ crosses
the x-axis 3 times and so has two imaginary roots.
By the complex conjugate root theorem,
the complex roots of $a(x)$ are conjugate pairs. Therefore $a(x)$ has three real roots, and two imaginary
roots. By the reasoning in the question above, there is a cycle of length 5 and a transposition between
the imaginary roots.
Thus the group for $a(x)$ is $S_5$ which is unsolvable implying there is no radical solution.
c
sage: a=x^5-4*x^4+2*x+2sage: ad=diff(a)
sage: sols=solve(ad, x)
sage: forsinsols:
....: print(s.rhs().n())
-0.259418669419159-0.411017935127584*I-0.259418669419159+0.411017935127584*I0.5311867963003053.18765054253801sage: plot(a, (-1, 4))
LaunchedpngviewerforGraphicsobjectconsistingof1graphicsprimitivesage: (sols[0].rhs() *sols[1].rhs()).n()
0.236233789039750-4.16333634234434e-17*Isage: (sols[0].rhs() *sols[1].rhs()).n().imag_part() <0.000000000001Truesage: # rounding error, so ignore that partsage: # 3 roots:sage: 0.236233789039750, 0.531186796300305, 3.18765054253801# for max and mins
(0.236233789039750, 0.531186796300305, 3.18765054253801)
We can see from the differentiated curve, that $a(x)$ is decreasing below $x = -1$ and increasing above $x = 4$.
Thus it crosses the x-axis three times, and so has three real roots, and two imaginary roots.
By the argument before this implies the group for this curve is $S_5$ which is unsolvable.
Q3
$$a(x) = (x - 2)^5 - (x - 2)$$
Let $a(x) = 0$ then
$$(x - 2)^4 = 1$$
The fourth roots of $1$ are $\pm i, \pm 1$. The remaining root of $a(x)$ is $x = 2$.
All these roots are real and solvable.
Q4
Substituting $y = x^2$, we get $a(x) = ay^4 + by^3 + cy^2 + dy + e$ which is easily solvable.
Any solution is then solvable for $x$ since $x = \pm \sqrt{y}$ which is itself a solvable equation.
Q5
There is no general solution for polynomials of degree 5, but there are polynomials of degree 5 which
have a solvable group.
B. Solvable Groups
Q1
Every subgroup of an abelian group is a normal subgroup.
Let $G$ be an abelian group with $x \in G$, and $H$ a subgroup with $a \in H$. Since $xax^{-1} = axx^{-1} = a \in H$, $H$ is a normal subgroup of $G$.
The set of commutators for an abelian group is ${ e } \implies Hxyx^{-1}y^{-1} = Hxy(yx)^{-1} = H \implies Hxy = Hyx \implies G/H$ is abelian.
From these two derivable properties of an abelian group, we see that every abelian group is also a solvable group.
Q2
The intersection of two subgroups of $G$ is a subgroup of $G$. For example $e \in J_0 = K \cap H_0$. For any $a \in J_0$, both $K$ and $H_0$ contain $a^{-1} \implies a^{-1} \in J_0$, likewise for products $a, b \in J_0 \implies ab \in J_0$.
All the iterated groups $J_i$ are subgroups of $K$, with $J_i \triangleleft J_{i + 1}$. Observe that $J_{i + 1}$ is a subgroup of $H_{i + 1}$. Let $x \in J_{i + 1}, a \in J_i$ then $xax^{-1} \in K \cap H_i = J_i$. Thus $J_i$ is a normal subgroup of $J_{i + 1}$.
Thus the sequence $J_0, \dots, J_n$ is a normal series of $K$.
Q3
$H_{i + 1} / H_i$ is abelian $\implies$$H_i$ contains all the commutators $xyx^{-1}y^{-1} \in H_{i + 1}$. Let $x, y \in J_{i + 1}$, then $xyx^{-1}y^{-1} \in J_{i + 1}$ and also $K$. Observe $xyx^{-1}y^{-1} \in H_i \cap K = J_i \implies J_{i + 1} / J_i$ is abelian. Thus the series ${ e } = J_0 \triangleleft J_1 \triangleleft \cdots \triangleleft J_n = K$ is a solvable series of $K$.
Q4
Combining the above two parts, we see that given a solvable group, any subgroup $K \subseteq G$ is also a solvable group.
Q5
$S_3$, the dihedral group of order 6 has six elements generated by $\langle a, b \rangle = { e, \alpha = a,
\beta = b, \delta = aba, \kappa = ab, \gamma = ba }$. The subgroup ${ e, \beta = b, \delta = aba }$$$\alpha^{-1} = \alpha \quad \beta^{-1} = \delta \quad \gamma^{-1} = \gamma$$$$\delta^{-1} = \beta \quad \kappa^{-1} = \kappa$$
\begin{alignat*}{2}
\alpha \beta \alpha^{-1} &= \alpha \beta \alpha &= \alpha \kappa &= \delta \
\kappa \beta \kappa^{-1} &= \kappa \beta \kappa &= \kappa \gamma &= \delta \
\gamma \beta \gamma^{-1} &= \gamma \beta \gamma &= \gamma \alpha &= \delta \
\alpha \delta \alpha^{-1} &= \alpha \delta \alpha &= \alpha \gamma &= \beta \
\kappa \delta \kappa^{-1} &= \kappa \delta \kappa &= \kappa \alpha &= \beta \
\gamma \delta \gamma^{-1} &= \gamma \delta \gamma &= \gamma \kappa &= \beta \
\end{alignat*}
So ${ \epsilon, \beta, \delta }$ absorbs products from $S_3$ and is a normal subgroup. Since $S_3$ has a solvable series, we conclude $S_3$ is a solvable group, and by part 4, that every subgroup is also solvable.
Q6
$$B = { e, (12)(34), (13)(24), (14)(23) }$$$$A_4 = { e, (12)(34), (13)(24), (14)(23), (13)(12), (12)(13), (14)(13), (13)(14), (14)(12), (12)(14), (24)(23), (23)(24) }$$$$[S_4: A_4] = 2$$
Every index 2 subgroup is abelian.
$B$ is the Klein subgroup of $A_4$ which is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. It is an abelian normal subgroup. Therefore also $A_4/B$ is abelian.
C. $p$th Roots of Elements in a Field
Q1
All roots of $x^p - a$ are of the form $d = \omega^k \sqrt[p]{a} : k \leq n - 1$. Since $\omega^{-k} \in F(\omega), \sqrt[p]{a} \in F(\omega, d)$.
See 31E5 and 31E6.
Q2
The question has a typo as explained here and should say "degree $\geq 2$".
$x^p - a$ is reducible in $F[x]$ by the question, so $x^p - a = p(x)f(x) = (x - z_1)(x - z_2) \cdots (x - z_p)$.
Since $x^p - a$ reduces to factors $p(x)$ and $f(x)$, and $z_i \notin F$, we conclude that $\deg p(x) \geq 2$.
Therefore $p(x) = (x - z_1)(x - z_2) \cdots (x - z_m)$ for some $m$ and $b = z_1 z_2 \cdots z_m \in F$. Likewise for $f(x)$.
Q3
From above $p(x)$ splits into linear terms of $p(x) = (x - z_1)\cdots(x - z_m)$ with a constant term $b = z_1 \cdots z_m$.
Since the roots of $x^p - a$ are of the form $\omega^j \sqrt[p]{a}$ so $b = (\omega^j \sqrt[p]{a})^m = \omega^{jm} \sqrt[p]{a}^m$.
But $d = \omega^{i} \sqrt[p]{a}$ or $\sqrt[p]{a} = \omega^{-i} d \implies b = \omega^{jm} (\omega^{-i} d)^m = \omega^k d^m$ for some $k$.
$c = b^s a^t$ is a solution for the equation $x^p - a$, so when $x^p - a$ is reducible it has a root in the field $F$.
Since $p(x) \in F[x]$ which means its constant term $b \in F$.
Otherwise we conclude that $F$ is irreducible over $F$.
D. Another Way of Defining Solvable Groups
Q1
By the definition, a subgroup is always contained in a maximal subgroup (if it's not maximal itself).
Because every finite group is a finite set, every chain of proper subgroups of a finite group has a maximal element and thus every finite group has a maximal subgroup. The same applies to maximal normal subgroups.
Q2
$J \triangleleft H$ and $\ran f = H$, so there exists a set $X$ of input values such that $f(X) = J$. Then $X = f^{-1}(J)$. For $a, b \in X, f(ab) = f(a)f(b) \in J$ which preserves group structure. Also $e_G \in J \implies e_H \in X$. And for the normal property let $g \in G$ then $f(g) f(a) (f(g))^{-1} = f(gag^{-1}) \in J \implies gag^{-1} \in X$.
Q3
Let $f: G \rightarrow G/K$ by $f(a) = Ka$. $\mathcal{F} \triangleleft G/K \implies f^{-1}(\mathcal{F}) \triangleleft G$. But $f^{-1}(\mathcal{F}) = \hat{\mathcal{F}}$.
Q4
This question is equivalent to proving the quotient group $G/K$ of a maximal normal subgroup $K$ is simple.
Let $H$ be a normal subgroup in $G/K$. Then $\hat{H}$ is the union of cosets in $H$. Then $K \triangleleft \hat{H} \triangleleft G \implies \hat{H} = G$.
Q5
Let $|G| = n = p_1 \cdots p_k$. Then for each $p_i$ there is an element $a \in G : \ord(a) = p_i \implies \langle a \rangle \subseteq G$. Thus there are $k$ subgroups in $G$ of order $p_i$.
$G$ has only the trivial subgroups ${ e }$ and $G \implies |G| = p$ for some prime $p$ with an element $a : \ord(a) = p$. Therefore $G = \langle p \rangle$.
$$|G/H| = 2 \implies H = H_0 \triangleleft H_q = G$$
and $H_{i + 1} / H_i$ is cyclic of order 2.
Now let $|G/H| = n$ and assume the statement is true for groups $|G/H| < n$.
If there is no subgroup $J$ between $H$ and $G$ such that $H \subseteq J \subseteq G$, then $H$ is a maximal normal subgroup of $G$. By part 4 above $G/H$ only contains trivial subgroups. By part 5, since the group is trivial it can only contain generators of the group which are prime order by Cauchy's theorem, and therefore $G/H$ is a cyclic group of prime order.
Lastly we deal with the case that $H$ is not a maximal normal subgroup where
$$H = H_0 \triangleleft H_1 \triangleleft \cdots \triangleleft H_{q - 1} = J \triangleleft H_q = G$$
Let $|H| = k$, then since $|G/H| = n$, $|G| = nk$.
$$H_0 \subseteq H_{q - 1} \implies |H_{q - 1}| > k \implies |H_q / H_{q - 1}| < n$$
Therefore by our inductive assumption, $H_q / H_{q + 1}$ is cyclic of prime order.
Likewise $H_1 \subseteq G \implies |H_1| < nk \implies |H_1/H_0| < n$. This can be generalized to
$$|H_i / H_0| < n \textrm{ for } i > 0$$
And $H_0 \subseteq H_i \implies |H_i| > k$$$\implies |G/H_i| < n$$$$\implies |H_{i + 1} / H_i| < n \textrm{ (by combining both statements) }$$
Which by our inductive assumption means that $H_{i + 1} / H_i$ is a cyclic group of prime order.
E. If $\Gal(K: F)$ Is Solvable, $K$ is a Radical Extension of $F$
To show $F_i$ forms an iterated normal extension, first we let $F_i = F_{i + 1}(c)$.
Assume $F_{i + 1}$ is a normal extension of $F_0$. $c^n = a \in F_{i + 1}$.
$$\pi(c) \pi(c) = \omega^2 c^2$$$$\implies \pi^k(c) = \omega^k c^k$$
\begin{align*}
\pi^p(c) &= \omega^p c^p \
&= c^p
\end{align*}
But $\Gal(F_i: F_{i + 1}) = \langle \pi \rangle$ so $\pi^p = e$ so $\pi^p(c) = c$$$c^p = c$$$$\pi^k(c) = \pi^k(c^p) = c^p$$
since $\Gal(F_i: F_{i + 1}) = \langle \pi \rangle$, and $\pi^k(c^p) = c^p$, all automorphisms fix $c^p$, and so $F_{i + 1}$ is the fixfield of $\Gal(F_i: F_{i + 1})$.
Q4
There are $p$ automorphisms $\pi^i \in \langle \pi \rangle$ which permute any root of $x^p - c^p$ to another unique root.
$x^p - c^p \in F_{i + 1}[x]$, and we know at least one root $b \in F_i$ so all roots are in $F_i$.
Q5
$F_q = F$ and $F_0 = K$, such that $F_q \subseteq \cdots \subseteq F_0$ such that each $F_i$ contains all roots of $x^p - c^p$ where $[F_i : F_{i + 1}] = p$. Thus each extension is radical over the previous one.
We conclude that $K$ is a radical extension of $F$.
