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Michael Artin Algebra, Chapter 2, 4.10
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| "name": "M. Artin, Algebra, Chapter 2, 4.10", | |
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| { | |
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| "source": [ | |
| "Для телеграм блога: https://t.me/VladimirsLoveForMath\n", | |
| "\n", | |
| "# Описание задачи:" | |
| ], | |
| "metadata": { | |
| "id": "GiucebiHLTd5" | |
| } | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "source": [ | |
| "" | |
| ], | |
| "metadata": { | |
| "id": "IU-1RtU0Irfo" | |
| } | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "source": [ | |
| "Приведите пример когда произведение двух элементов имеющих конечный порядок необязательно порождает элемент имеющий конечный порядок. Что если эта группа является абелевой?\n", | |
| "\n", | |
| "Другими словами:\n", | |
| "\n", | |
| "Пусть $\\exists x, y \\in G, \\exists n, m \\in \\mathbb{N}\\setminus\\{0\\}$ такие что: $x^n = 1$ и $y^m = 1$. Т.е. элементы $x$ и $y$ образуют две циклические группы порядков $n$ и $m$ соответственно.\n", | |
| "\n", | |
| "Приведите пример таких $x$ и $y$ чтобы $(xy)^k \\neq 1$ для любых $k \\in \\mathbb{N}\\setminus\\{0\\}$." | |
| ], | |
| "metadata": { | |
| "id": "WFzXBao0I4XN" | |
| } | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "source": [ | |
| "# Случай когда группа $G$ абелева\n", | |
| "\n", | |
| "Легче начать с конца, рассмотреть случай когда группа $G$ является абелевой.\n", | |
| "\n", | |
| "Абелева, значит произведение будет коммутативным, т.е. $xy = yx$. Тогда:\n", | |
| "\n", | |
| "$(xy)^k = (xy\\cdot xy \\dots ) = (xx \\dots yy \\dots) = x^k y^k$\n", | |
| "\n", | |
| "Мы так можем сделать только если произведение коммутативно.\n", | |
| "\n", | |
| "Возьмём $k = nm$, тогда:\n", | |
| "\n", | |
| "$(xy)^k = x^k y^k = x^{nm} y^{nm} = (x^n)^m (y^m)^n = 1^m 1^n = 1$\n", | |
| "\n", | |
| "Соответственно, мы доказали что для абелевой группы такая ситуация невозможна. Всегда найдётся $k = nm$ для которого $xy$ будет порождать циклическую группу\n" | |
| ], | |
| "metadata": { | |
| "id": "h0RHW_ziLjU7" | |
| } | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "source": [ | |
| "# Основной случай\n", | |
| "\n", | |
| "Итак, группа $G$ не является абелевой и произведение в общем случае некоммутативно. Яcно что мы не можем рассматривать числовые множества ($\\mathbb{Z}$, $\\mathbb{R}$, $\\mathbb{C}$) на роль $G$ так все они являются абелевыми.\n", | |
| "\n", | |
| "Самое простое что нам доступно, где умножение некоммутативно, это матрицы.\n", | |
| "\n", | |
| "Пусть $G = GL_2(\\mathbb{R})$, т.е. группа обратимых (невырожденных) $2 \\times 2$ матриц. Возьмём $n = m = 2$ и подберём такие матрицы $X, Y\\in G$ что:\n", | |
| "\n", | |
| "$X^2 = I$\n", | |
| "\n", | |
| "$Y^2 = I$\n", | |
| "\n", | |
| "Буквально наугад подобрал такие матрицы:\n", | |
| "\n", | |
| "$\n", | |
| "X = \\begin{bmatrix}\n", | |
| "-1 & a \\\\\n", | |
| "0 & 1\n", | |
| "\\end{bmatrix}\\quad\n", | |
| "Y = \\begin{bmatrix}\n", | |
| "-1 & b \\\\\n", | |
| "0 & 1\n", | |
| "\\end{bmatrix}\n", | |
| "$\n", | |
| "\n", | |
| "Окей, смотрим что за матрицу они порождают:\n", | |
| "\n", | |
| "$XY = \\begin{bmatrix}\n", | |
| "1 & a-b \\\\\n", | |
| "0 & 1\n", | |
| "\\end{bmatrix}$\n", | |
| "\n", | |
| "Эта матрица кое-что напоминает. В первой главе было задание 1.6, откуда мы вывели следующее:\n", | |
| "\n", | |
| "$\\begin{bmatrix}\n", | |
| "1 & x \\\\\n", | |
| "0 & 1\n", | |
| "\\end{bmatrix} ^ n = \\begin{bmatrix}\n", | |
| "1 & nx \\\\\n", | |
| "0 & 1\n", | |
| "\\end{bmatrix}$\n", | |
| "\n", | |
| "Матрица $XY$ как раз имеет именно такой вид. Получается что:\n", | |
| "\n", | |
| "$(XY)^n = \\begin{bmatrix}\n", | |
| "1 & n(a-b) \\\\\n", | |
| "0 & 1\n", | |
| "\\end{bmatrix}$\n", | |
| "\n", | |
| "Нам осталось убедиться что $(XY)^k \\neq I$ для любого $k$.\n", | |
| "\n", | |
| "Т.е. нам нужно гарантировать что $k(a-b) \\neq 0$. Так как $k\\neq 0$, то $a \\neq b$.\n", | |
| "\n", | |
| "Мы можем взять $a=1$, $b=2$ и всё должно работать." | |
| ], | |
| "metadata": { | |
| "id": "OudP4VBYOtNz" | |
| } | |
| }, | |
| { | |
| "cell_type": "markdown", | |
| "source": [ | |
| "# Проверка" | |
| ], | |
| "metadata": { | |
| "id": "8R8MdVqrWdFi" | |
| } | |
| }, | |
| { | |
| "cell_type": "code", | |
| "source": [ | |
| "import numpy as np\n", | |
| "\n", | |
| "a = 1\n", | |
| "b = 2\n", | |
| "I = np.eye(2)\n", | |
| "X = np.array([\n", | |
| " [-1, a],\n", | |
| " [0, 1]\n", | |
| "])\n", | |
| "Y = np.array([\n", | |
| " [-1, b],\n", | |
| " [0, 1],\n", | |
| "])\n", | |
| "\n", | |
| "# check that indeed x^2 = 1, y^2 = 1\n", | |
| "assert np.array_equal(X@X, I)\n", | |
| "assert np.array_equal(Y@Y, I)\n", | |
| "\n", | |
| "XY = X@Y\n", | |
| "kth_XY = XY\n", | |
| "# try to multiply by itself\n", | |
| "for k in range(1,50):\n", | |
| " print(kth_XY)\n", | |
| " kth_XY = kth_XY@XY\n", | |
| " # we should NOT get I for any k\n", | |
| " assert not np.array_equal(kth_XY, I)\n" | |
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