My attempt:  Consider the map $σ : F→ F$ such that $∀a ∈ F, σ(a) = a^p$. Then $σ$ is a field automorphism of $F$.  $ker(σ) = 0$ and since $F$ is a finite field, $σ ∈ Gal(F : K)$, the galois group of order $k$. Since the order of the galois group is $k$ it is generated by $σ$ so we can deduce that $Gal(F : K)$ is a cyclic group $Z_n$.  By the Fundamental Theorem of Galois Theory,  $|G| = |Gal(F : K)| = [F : K]$, for the field extension  $F : K$ of finite degree.  $Gal(F : K)$ is a cyclic group of order $[F : K]$.
13733
I have to compute Galois group of $\mathbb F_q/\mathbb F_p$ where $q=p^n$. I know already that $\mathbb F_q/\mathbb F_p$ is galois, so I don't need to prove it. Moreover, I know that \begin{align*}\sigma :\mathbb F_q&amp;\longrightarrow \mathbb F_q\\ x&amp;\longmapsto x^p\end{align*} is an automorphism that fix elements of $\mathbb F_p$ and thus $\sigma \in Gal(\mathbb F_q/\mathbb F_p)$.  Now, I know that if $k&lt;p^n$ then $\sigma ^k(x)=(x^{p})^k=x^{pk}\neq 1$ and thus $\sigma $ is of order $p^n$. Now if I can prove that $|Gal(\mathbb F_q/\mathbb F_p)|=p^n$, then $Gal(\mathbb F_q/\mathbb F_p)=\left&lt;\sigma \right&gt;$ and it would be finish, but I can't prove it since even $X^q-X$ split over $\mathbb F_q$, theis polynomial is not irreducible, and thus, I can't conclude. So how can I find an irreducible polynomial of degree $q$ that split over $\mathbb F_q$ ?
100392
I am attempting to prove what the title says, that $\mathbb{F}_p/\langle f(x)\rangle$ with $f(x)$ irreducible of degree $n$ is a field with $p^n$ elements.  I have already proven that for any field $K$ and polynomial $f(x)$ in $K[x]$, $K[x]/\langle f(x)\rangle$ is a field if and only if $f(x)$ is irreducible in $K[x]$. So I know that $\mathbb{F}_p/\langle f(x)\rangle$ is a field for sure. Where I'm lost is figuring out how to know for sure that there are $p^n$ elements. It's fairly clear if $n\in\{ 0 ,1\}$, but I don't think induction is going to help.  I know from a proposition in class that since $F_p = \mathbb{Z}/\langle p\rangle$ and $F_p = \mathbb{Z}/\langle p\rangle$ is an integral domain, $F_p[x] = (\mathbb{Z}/\langle p\rangle)[x] = (\mathbb{Z}[x])/(\langle p\rangle[x])$, but I'm not sure whether or not that's helpful information.
20021
I have a finite field $\mathbb{F}_{27}$. I need to find an element of order 13. I know that multiplicative group of this field is cyclic with order 26. So I want to find a generator $g$ of this cyclic group, then $g^2$ will have order 13. Also I know that all elements of this field can be represented by polynomials from $\mathbb{F}_3[t]$, where t is root of irreducible polynom $x^3-x-1$.  Probably we can find generator of $\mathbb{Z}_{26}$ and than build an isomorphism to $\mathbb{F}_{27}\setminus 0$.  Anyway I don't know were to start.  Thanks!
93872
If $G$ is a finite group and $T$ an automorphism of $G$ such that $T$   sends more than $\frac{3}{4}|G|$ elements to their inverse. Then we have show that $$T_x=x^{-1}$$ for all $ x \in G$ and $G$ is abelian.   I construct $ S= \{  x \in G : T_x=x^{ -1}  \}$ and for $ b \in G $, $ S_b= \{ xb:x\in S \}$ then $| S \cup S_b|\geq (1/2)|G|$. Let $ a \in S\cap S_b $ then $ a=tb $ for some $t\in S $. I dont know how to solve it by this construction or any other method. Give me any kind of help.
8290
Assume that $\pi$ is an automorphism of a finite group $G$. Let $S$ denote the set $\lbrace g \in G: \pi (g)=g^{-1} \rbrace$. Show that if $|S|&gt;\frac{3|G|}{4}$, then $G$ is an abelian group.    Anyone has any idea on how to solve this ? I have no idea how to start.
91654
Let $K$ be a field and $f \in K[x]$ a polynomial of degree $n&gt;0$. Assume that $E$ is an extension of $K$ such that $f$ can be expressed as a product of linear factors over $E$:  $$f(x)=\gamma (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n).$$  I want to show that $\lvert K(\alpha_1,\dots,\alpha_2) : K \rvert$ is a divisor of $n!$. I also want to show that if $\lvert K(\alpha_1,\dots,\alpha_n): K \geq (n-1)!$ then $f$ is irreducible over $K$ or $f$ already has a root in $K$.  The problem has a hint. It says: "Use induction on $n=\deg f$. If $f=gh$ in $K[x]$ with $\deg g =r$ and $\deg h=n-r$, then $r!(n-r)!$ is a divisor of $n!$.  Any idea for solving this problem would be appreciated.
104898
Let $f: G \to H$ be a homomorphism of groups. Assume that $a\in G$ and $\operatorname{ord}(a)=n$. Prove that the order of $f(a)$ is a divisor of $n$. I know that if $H$ is a subgroup of $G$, then $$|G|=|H|\cdot|\text{distinct cosets of $H$}|$$ so $|H|$ must be divisor of $|G|$, but I don't understand how to apply that logic to homomorphisms (or if that will even work). Any help is appreciated!
37717
I want to show without using Sylow theorem that     Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)   My attempt: Since $|G|=2p,$ even $\exists~a\neq e\in G$ such that $a^{-1}=a.$ Then $H=\{e,a\}\leq G.$ Let $S$ be the set of all left cosets of $H$ in $G.$ Now $S$ forms a group w.r.t. the composition $g_1Hg_2H=(g_1g_2)H.$ The mapping $\sigma:S\to G:gH\mapsto g$ is an $1-1$ homomorphism. Then $S\simeq\sigma(S)\leq G.$ Thus $G$ has a subgroup of order $p.$  Please tell me whether I'm right!     I'm skecptical as I didn't use that $p$ is prime.
27428
Let $K$ be a field and $f \in K[X]$ of degree $n$ with Galois group $S_n$. Let $a$ be a root of $f$, $L = K(a)$, and let $E$ be a intermediate field of the extension $K \subset L$. Prove that $E = K$ or $E = L$.   Any suggestions on where to start? Thanks.
87422
The book "A First Course in Abstract Algebra" by Fraleigh says      If $G$ is a finite subgroup of the multiplicative group $\langle F^*,\cdot\rangle$ of a field $F$, then $G$ is cyclic. In particular, the multiplicative group of all nonzero elements of a finite field is cyclic.    I wonder why that is. Take $a,b\in F$ such that $a^2=b^2=1$. Then $\{1,a,b,ab\}$ is a non-cyclic subgroup of $F$. Where am I going wrong?  Thanks in advance!
52539
In Grove's book Algebra, Proposition 3.7 at page 94 is the following     If $G$ is a finite subgroup of the multiplicative group $F^*$ of a field $F$,   then $G$ is cyclic.   He starts the proof by saying "Since $G$ is the direct product of its Sylow subgroups ...". But this is only true if the Sylow subgroups of $G$ are all normal. How do we know this?
43262
Let $G$ be a cyclic group of order $m$ generated by an element $a$.  I want to show that the order of $a^k$ is $m/d$, where $d:=\gcd(k,m)$.  I have a simple proof, but want to make sure I haven't overlooked anything since the other proofs I've seen (such as in Dummit and Foote) look more complicated.    The order of $a^k$ is the cardinality of the set $\{a^{ks}: s \in \mathbb{Z}\}$, which equals the cardinality of the set $\{a^{ks} a^{mt}: s,t \in \mathbb{Z} \}$ since $a^m$ is the identity.  But the set $\{ks+mt: s,t \in \mathbb{Z} \}$ is the set $d \mathbb{Z}$ by the basic properties of integers. So we have that the cardinality of the set is $| \langle a^k \rangle| = |\{1, a^d, a^{2d}, \ldots, a^{m-d} \}| = m/d$.
118804
I am having difficulty with this proof. I have previously proved that for an automorphism on the extension field $K$ of $F$ that when restricted to $F$ is the identity automorphism must take roots of $f(x)\in F[x]$ in $E$ to roots of $f(x)$ in $E$. I was thinking of using the minimal polynomial $x^{3}-2\in\mathbb{Q}[x]$ and the result of previous problem I solved, but I suspect there are automorphisms that don't fix the subfield, so the result can't be used or I need to prove that any automorphism of an extension field must fix the subfield that is being extended or always sends roots in $E$ to roots in $E$.  I am not asking for a solution to this, but maybe a hint or pointer in the right direction. This problem is introduced in the text prior to Galois theory, so please keep that in consideration.
29009
I want to prove the following:          Let $G$ be a group of order $2^nm$, where $m$ is odd, having a cyclic Sylow $2$-subgroup.     Then $G$ has a normal subgroup of order $m$.      ATTEMPT:  We will show that $G$ has a subgroup of order $m$. Let $\theta:G\to \text{Sym}(G)$ be the homomorphism which is defined as $$  \theta(g)=t_g,  \quad  \forall g\in G $$ where $t_g:G\to G$ is defined as: $$  t_g(x)=gx,  \quad  \forall x\in G. $$ Let $g$ be an element of order $2^n$ in $G$. (There exists such an element since $G$ has a cyclic Sylow $2$-subgroup.)  The cyclce representation of $t_g$ is a product of $m$ disjoint cycles each of length $2^n$. Therefore $t_g$ is an odd permutation.  Thus the homomorphism $\epsilon\circ \theta:G\to\{\pm 1\}$, where $\epsilon:\text{Sym}\to\{\pm 1\}$ is the sign homomorphism, is a surjection.  By the First Isomprphism Theorem, we conclude that the kernel $K$ of $\epsilon\circ \theta$ is of order $2^{n-1}m$.  If $n$ were equal to $1$ then we are done.  If $n&gt;1$ then note that any Sylow $2$-subgroup of $K$ is also cyclic. This is because each Sylow $2$-subgroup of $K$ is contained in a Sylow $2$-subgroup of $G$, where the latter is cyclic.  Now we can inductively show that $K$ has a subgroup of order $m$.    What I am struggling with is showing the normality.  Can anybody please help me with this.  Thanks.
105855
After a week in my Abstract Algebra class, the professor proposed this as a problem. I'm not entirely sure where to begin. $ V = \{ e, \tau, \tau_1, \tau_2 \}$, so I'm not sure exactly what is meant by $\text{Aut} (V)$. Is it simply saying that the only way to have an automorphism on $V$ is to rearrange the order of the elements of $V$ and therefore $\text{Aut} (V) \cong S_3$?
45977
Suppose $K$ is a prime subfield of $E$, then if $\phi$ is an automorphism from $E$ to $E$, we have for all $x \in K$, $\phi(x) = x$.   I feel like this is just the definition of a field automorphism, but my book says this should be proven as an exercise.
17435
Suppose $K$ is an extension of $\mathbb{Q}$ in $\mathbb{C}$, where $Gal(K/\mathbb{Q})$ is cyclic of order 4. Show that $i\notin K$.   ($i$ is the imaginary number $i^2=-1$.)  My Galois theory is quite weak, hope someone can check if my attempt is correct.    My attempt:  Suppose to the contrary $i\in K$. Let $\sigma\in Gal(K/\mathbb{Q})$.  Note that $\sigma (i)\sigma(i)=\sigma(i^2)=\sigma(-1)=-1$ since $\sigma$ fixes $\mathbb{Q}$.  This means that $\sigma(i)=i$ or $\sigma(i)=-i$. The first case is ruled out since $\sigma$ only fixes $\mathbb{Q}$.  So $\sigma(i)=-i$. This means that $\sigma(a+bi)=a-bi$, so $\sigma$ is effectively complex conjugation, which has order 2.   Since $\sigma$ was arbitrary, this contradicts that $Gal(K/\mathbb{Q})$ has an element of order 4.    Is this ok?  Thanks.  Update: Now I see that my argument is clearly flawed. What would be the correct proof?