Let $F$ be a field and let $G$ be a finite subset of $F \setminus \{0\}$ containing $1$ and satisfying the condition that if $a, b ∈ G$ then $ab^{−1} ∈ G$. Show that there exists an element $c ∈ G$ such that $G = \{c^i : i ≥ 0\}$.  This is an excercise from the book: The Linear Algebra a Beginning Graduate Student Should Know; Golan
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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?
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I need help with this:  If a is an element of $\mathbb Z_n$ and $\gcd(a,n) &gt; 1$ , then a is not invertible.   First you show that if $a$ is an element of $\mathbb Z_n$ and $\gcd(a,n)&gt;1$, then there  is an element $b$ of $\mathbb Z_n$ and ($b$ is not equal to the zero element) from which $ab = 0$.   The second part I cant get. i.e. show if $b$ is not the zero element  and $ab=0$ then $a$ is not invertible.
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Well be with you, gentlemen. I have the following problem from Aluffi's Algebra:  given a finite group $G$ with an unique element $f$ of order $2$, show that \begin{equation} \prod_{g\in G}g=f \end{equation} My reasoning is the following. Since $f$ is the unique element of order $2$, for each $g\in G$ such that $g\neq e$ and $g\neq f$, it must be that $g^{-1}\neq g$. So we can concoct a particular product coupling the elements with their inverses: \begin{equation} e\cdot f\cdot (g_1\cdot g_1^{-1})\cdot(g_2\cdot g_2^{-1})\cdots (g_n\cdot g_n^{-1}) \end{equation} which is really just $f$ in the end. My question is: how can I show that the order of the factors in this product really does not matter? I cannot see a way out.
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I am confused on how to get this problem started. I want to say that they are associated, but I am not sure if that is correct.   Let $G$ be the subset of $F[x]$, $F$ a field, consisting of all the polynomials whose constant term is nonzero.      Show that $a_0+a_1x+\cdots+a_nx^n$ is irreducible iff $a_n+a_{n-1}x+\cdots+a_0x^n$ is irreducible.    Like I said, I am not sure how to get started on this... A hint would be great, with a hidden answer would be much appreciated. I haven't been able to think of a way to get started on this.
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Suppose $a$ and $b$ are elements of a finite field of order $2^n$ with $n$ odd and $a^2+ab+b^2=0$. Is it necessary that both $a$ and $b$ must be zero ?  I understand that the field has characteristic $2$ but don't know how to use the fact that $n$ is odd, please help.
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I need help with this proof:     Let $A$ and $B$ be sets. Prove that $A\cap (B\setminus A)\subseteq \varnothing$.   My problem: Since $B\setminus A$ = $B\cap A^{c}$, we can say that if $x \in B$ then $x\in B\cap A^{c}$and $x \in A^{c}$ as well. So $x\notin A$.  Where do I go from here?
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Take $G$ to be a group of order $600$. Prove that for any element $a$ $\in$ G there exist an element $b$ $\in$ G such that $a = b^7$.    My thought process: since $a = b^7$ $\implies$ $|a| = |b^7|$. Consequently   $\operatorname{lcm}(1,|a|) = \dfrac{1}{7}\operatorname{lcm}(7,|b|)$ implies  $7|a| = 7|b|$ so $|a| = |b|$. I don't know where I would go from here or if this is even the right approach.
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How would you prove that 2 elements of the group $G$ are equal, i.e., $a = b$ if we know that $a$, $b$ belong to group $G$ and that $m$, $n$ are relatively prime (integers) and that $a^m = b^m$ and $a^n = b^n$  Thanks!
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Let $G$ be a finite and non-abelian group.  How do I prove the following statement? $$\exists a,g,h\in G \colon\quad h=aga^{-1},\ g\neq h ,\ gh=hg.$$  Thanks in advance.
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Lets $A\subset \mathbb F$ where $\mathbb F$ is an arbitrary field. Is is correct, that there are two possibilities?   $A=\{0\}$, then $\gcd(A)=A$ $A\neq\{0\}$, then $\gcd(A)=A\backslash\{0\}$   My reasoning here would be: If $0$ is a common divisor of all elements in $A$, then $\gcd$ must be divisible by $0$ which is only the case for $0$ itself. If any nonzereo element is in $A$, then a common divisor can only be another nonzero element. This yields $A\backslash\{0\}$ because those are all associated.
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Problem: Show that if $(b,c)=1$, then for any integer $a$ we have $(a,bc)=(a,b)(a,c)$. [Hint: Prove that each member of the alleged equation divides the other.]  This is the way I proved it (without the hint): $(a,b)(a,c)=(a(a,b),c(a,b))=(a^2,ab,ac,bc)=(a^2,bc,a(b,c))=(a^2,a,bc)=((a^2,a),bc)=(a,bc)$  But how should I prove it using that hint or at least a similar proof.
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For which $c$ will the set $K(c) = \{a + b\sqrt{c} : a, b \in \mathbb{Q}\}$ be a field? I know for example, that $K(\frac{2}{3})$ will be one, I am just wondering what the most general result of this type is.
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Let $(B, +, \cdot)$ be a non-trivial ring with the property that every $x \in B$ satisfies $x \cdot x = x$. How does one prove that such a ring $(B, +, \cdot)$ must have a unit element $1_B$? (Or, in case this is not true in general, what is a counterexample?)   BTW, I'm looking for an elementary proof, not requiring anything more than the definition of a ring, the definition of $(B, +, \cdot)$, and, if necessary, the easily shown facts that $x + x = 0$ and $x\cdot y = y\cdot x,\,\forall\, x,y \in B$.
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Here is a proof I saw somewhere of the fact $R/I$ is a field if and only if $I$ is maximal:  $\implies$ Suppose that $R/I$ is a field and $B$ is an ideal of $R$ that properly contains $I$. Let $b \in B$ but $b \notin I$. Then $b + I$ is a nonzero element of $R/I$ and therefore there exists an element $c + I$ such that $(c + I)(b + I) = 1 + I$. Since $b \in B$ we have $bc \in B$. Because $1 + I = (c + I)(b + I) = bc + I$ we have $1 - bc \in I \subset B$. So $1 = (1-bc) + bc \in B$. Hence $B = R$.  $\Longleftarrow$ Now suppose $I$ is maximal and let $b \in R$ but $b \notin I$. Consider $B = \{br + a \mid r \in R, a \in I \}$. This is an ideal properly containing $I$. Since $I$ is maximal, $B = R$. Thus $1 = bc + a^\prime$ for some $a^\prime \in I$. Then $1 + I = bc + a^\prime + I = bc + I = (b + I)(c + I)$.  I thought this was fairly long so I tried to come up with a shorter proof. Can you tell me if this is right:  $\implies$ Assume that $R/I$ is a field and $I$ is not maximal. Then there exists an $x \in R - I = I^c$ that is not a unit (otherwise $I$ would be maximal). Then $x + I$ does not have an inverse hence $R/I$ is not a field.  $\Longleftarrow$ Assume $I$ is maximal and $R/I$ is not a field. Then there is an $x$ such that $x + I \neq 0 + I$ does not have an inverse. This $x$ is not in $I$ and $x$ is not a unit. Hence $I \subsetneq I + (x) \subsetneq R$. Which contradicts $I$ being maximal.
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More homework help.  Given the function $f:A \to B$.  Let $C$ be a subset of $A$ and let $D$ be a subset of $B$.  Prove that:  $C$ is a subset of $f^{-1}[f(C)]$  So I have to show that every element of $C$ is in the set $f^{-1}[f(C)]$  I know that $f(C)$ is the image of $C$ in $B$ and that $f^{-1}[f(C)]$ is the pre-image of $f(C)$ into $A$.  Where I'm stuck is how to use all of this information to show/prove that $C$ is indeed a subset.    Do I start with an arbitrary element (hey, let's call it $x$) of $C$? and then show that $f^{-1}[f(x)]$ is $x$?  I could use a little direction here...  Thanks.
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If $A\subseteq B$ under what conditions is $f^{-1}(A) \subseteq f^{-1}(B)$, where $f^{-1}$ is the preimage, not  the inverse.
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I will quote a question from my textbook, to prevent misinterpretation:     Let $G$ be a finite abelian group and let $m$ be the least common multiple of the orders of its elements. Prove that $G$ contains an element of order $m$.   I figured that, if $|G|=n$, then I should interpret the part with the least common multiple as $lcm(|x_1|,\dots,|x_n|)=m$, where $x_i\in G$ for $0\leq i\leq n$, thus, for all such $x_i$, $\exists a_i\in\mathbb{N}$ such that $m=|x_i|a_i$. I guess I should use the fact that $|x_i|$ divides $|G|$, so $\exists k\in \mathbb{N}$ such that $|G|=k|x_i|$ for all $x_i\in G$. I'm not really sure how to go from here, in particular how I should use the fact that $G$ is abelian.