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4

Yes. This paper " V. V. Pylaev, N. F. Kuzennyi, Finite groups with a cyclic maximal subgroup, Ukrainian Mathematical Journal, 28 (5) (1976), 500--506", classify finite groups with a cyclic maximal subgroup.


3

Let $G$ be the group in question. In $G$, we have: $x^2 = y^2x^2y$ (1), and $yx^{-1}y^2=x^7$ (2) $\Rightarrow y^2=xy^{-1}x^7$ Substitute in (1) $x^2=xy^{-1}x^7x^2y \Rightarrow yxy^{-1} = x^9$ (3) (3) $\Rightarrow yx^{-1} = x^{-9}y$, substitute in (2) to get $x^{-9}y^3 =x^7\Rightarrow y^3 = x^{16}$. Since the element $y^3=x^{16}$ commutes with both ...


3

I guess you are possibly making the common mistake assuming the Galois fields $\operatorname{GF}(n)$ are the quotient rings $\mathbf{Z}/n\mathbf{Z}$, they aren't! (There are several questions on this site you can find explaining what they actually are in more detail). In fact the Galois field $\operatorname{GF}(8)$ can be explicitly constructed as $$ ...


2

Let $xS^k \in S/S^k$ and $yS^k \in S^{k-1}/S^k$ with $x \in S$, $y \in S^{k-1}$. To show that $S^{k-1}/S^k \le Z(S/S^k)$, we have to show that $xS^k$ and $yS^k$ commute with each other or, in other words, that their commutator is the trivial element of $S/S^k$. Now $[xS^k, yS^k] = [x,y]S^k$. But, since $x \in S$ and $y \in S^{k-1}$, $[x,y] \in [S,S^{k-1}] = ...


2

$$x^2=y^2x^2y$$ $$x=x^{-1}y^2x^2y=y^{-1}(yx^{-1}y^2)x^2y$$ $$x=y^{-1}x^9y$$ then we can say that; $$y^2x^2y=x^2=y^{-1}x^{18}y$$ $$y^2x^2=y^{-1}x^{18}$$ $$y^3=x^{16}$$ Now, $$(xy^2)^2=yx^2$$ $$xy^2xy^2=yx^2$$ $$xy^2x(y^3)=yx^2y=y^3x^2y^2=x^{18}y^2$$ $$y^2x^{17}=x^{17}y^2$$ $$y^5x=xy^{5}$$ As a last step; $$x^2=y^2x^2y$$ ...


2

First we manipulate the relations to show that the group $G$ defined by the presentation is the direct product $\langle x\rangle\times\langle y\rangle$, with $x$ of order a divisor of $8$ and $y$ of order a divisor of $3$. Then, to show that $x$ and $y$ have orders $8$ and $3$ exactly, we exhibit a homomorphism from $G$ onto a group in which the images of ...


1

I computed the powers of $z = xy$: $xy$, $y^2$, $x$, $y$, $xy^2$, $x^2$, $xy$, $y^2$, $1$. This in particular gives $z^7 = z$ so $z^6 = x^2 = 1$, and $x^3 = x^2 = 1$ gives $x = 1$. All you need from this though is $z^2 = y^2$ $z^3 = x$ So $1 = y^6 = (z^2)^3 = (z^3)^2 = x^2$, then the last step of $x^3 = x^2 = 1 $ implies $x = 1$. I posted this ...


1

GF(8), the field with 8 elements is definitely not $\mathbb{Z}_8$. One way to realize GF(8) is as $\mathbb{Z}_2[\alpha]$ where $\alpha$ satisfies the relation $\alpha^3=\alpha+1$ (all arithmetic is done mod 2. Equivalently GF(8) may be thought of as $\mathbb{Z}_2[x]/<x^3+x+1>$ (quotient ring). In the first realization we would have ...


1

Most typically we have $GF(8)=\Bbb{F}_2[x]/\langle x^3+x+1\rangle$, and we use shorthand notation $q(2)$ for the element $q(x)+\langle x^3+x+1\rangle$. For that to work we first interpret the coefficients of $q$ from $\Bbb{F}_2=\{\overline{0},\overline{1}\}$ as integers $0,1$, and then we can evaluate $q(2)$ as an integer. Let us denote $\alpha=x+\langle ...


1

Rats, I misread the original question. You're looking for $\text{Aut}(\Bbb Z/5\Bbb Z)$, which is indeed isomorphic to $\Bbb Z/4\Bbb Z$. Since this group is cyclic, it's enough to find an element of order $4$, and then all choices are given by powers of that generator. Indeed, multiplication by $-1$ gives you inversion, which is an element of ...



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