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Conjugation amounts to changing perspective. Here are three major examples to illustrate this point: notational representations of permutations with respect to labels on objects, matrices representing linear transformations with respect to ordered bases, and loops a la the fundamental group as studied in algebraic topology and homotopy theory. $\sf ... 4 It follows that this is false from the same result about algebraic numbers. Every algebraic number can be written as$\frac{\alpha}n$where$n$in an integer and$\alpha$is an algebraic integer. So if this was true for algebraic integers, it would be also true for algebraic numbers. The fact that there are algebraic numbers that cannot be expressed in ... 3 Yes! If$L/F$is abelian then any intermediate extension$K/F$corresponds to a normal subgroup and hence is also Galois. In particular, cyclotomic extensions are abelian. 3 Since an element always commutes with its powers, we have that every element has order 2. In particular, this implies$a = a^{-1}$for every$a \in G$. Also, a group with this property must be abelian, since$[a, b] = aba^{-1}b^{-1} = abab = (ab)^2 = e$. Now, if the group has more than two elements, all these should commute, which contradicts the initial ... 2 What you're looking for is a Cohen ring. The proof that Cohen rings exist can be found here: http://stacks.math.columbia.edu/tag/0323 2 Let's look look what happens to each element of$\{1,2,3,\dots,k\}$under the permutation:$\pi(a_1\ a_2\ \dots\ a_n)\pi^{-1}$(here, of course, we must have$n \leq k$). Let's denote$\pi(a_i) = b_i, i = 1,2,\dots,n$. Suppose$j \not\in \{b_1,b_2,\dots,b_n\}$. Then, since$\pi$is bijective,$\pi^{-1}(j) \not\in \{a_1,a_2,\dots,a_n\}$. Thus$(a_1\ ...

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In the first case, they mean a composition of permutations. In the second case, they are thinking of $\pi$ as a function as well. Consider $\pi \in S_5$ where $\pi = (1~3)$. Then $\pi(1) = 3, \pi(2) = 2, \pi(3) = 1, \pi(4) = 4$ and $\pi(5) = 5$. Now, lets see if the problem works in your example: on the one hand, $$\pi(4~5~3)\pi^{-1} = (1~3)(4~5~3)(1~3) ... 2 Denote by A_g the element of P that contains g \in G. The proof mostly uses the facts that the operation is well-defined and A_g=A_1 \iff g \in A_1 Let's check that A_1, the element of P containing the identity, is a subgroup: Identity: It's given that 1 \in A_1 Closure: Suppose a, b \in A_1. Then A_a = A_1 and A_b=A_1. By the ... 2 You can choose a permutation a (others will work, too) such that: a(1) = 5, a(2) = 6, a(3) = 1 and a(4) = 3. It doesn't matter what we choose for a(5),a(6), as long as we don't pick from the set \{1,3,5,6\}, since those values are already "taken". a(5) = 2, and a(6) = 4 will do. Thus a = (1\ 5\ 2\ 6\ 4\ 3) is one possibility. 2 This is another conjugation problem in disguise: (2\ 3)(3\ 4)(4\ 5)(4\ 3)(3\ 2) = (2\ 3)[(3\ 4)(4\ 5)(3\ 4)^{-1}](3\ 2) = (2\ 3)(3\ 5)(2\ 3)^{-1} (since (3\ 4) takes 4 \to 3 and fixes 5) = (2\ 5) (since (2\ 3) takes 3 \to 2 and fixes 5). 2 Cyclic groups to the rescue! If an infinite group has an element of infinite order (that is, a subgroup isomorphic to \Bbb Z) then it has an infinite number of subgroups (because \Bbb Z does). Otherwise, every element is of finite order, in which case, if we only have a finite number of subgroups G = \bigcup\limits_k \langle g_{i_k}\rangle, which is ... 2 Say G is infinite. Let S(x) denote the subgroup generated by x\in G. If there exists x such that S(x) is infinite then G has infinitely many subgroups, since an infinite cyclic group has infinitely many subgroups. On the other hand if every S(x) is finite then there must be infinitely many distinct subgroups of the form S(x), since ... 2 If N is finite, then every element of N has finite order. Since g^n \in N, g^n has finite order, thus: (g^n)^k = e for some k > 0, that is: g^{kn} = e, contradicting that g has infinite order. P.S.: you already used the fact that N is normal, when you said (gN)^n = g^nN, which only holds if coset multiplication does-that is, if N ... 1 Let's go through the product(s)$$ (2 3) (3 4) (4 5) (4 3) (3 2) $$step by step. Start with 2, by writing (2, and see how the elements map out: 2\mapsto 3\mapsto 4\mapsto 5\mapsto 5\mapsto 5\qquad:\qquad(2\; 5 5\mapsto 5\mapsto 5\mapsto 4\mapsto 3\mapsto 2\qquad:\qquad(2\; 5)(3 3\mapsto 2\mapsto 2\mapsto 2\mapsto 2\mapsto 3\qquad:\qquad(2\; ... 1 Your professor is saying the following. The left-action of G on itself gives a "permutation representation,'' that is, an embedding G \hookrightarrow S_{2k}. If you are not familiar with the phrase "action" in the previous sentence, all I am doing is copying the proof of Cayley's theorem that every group is a subgroup of the symmetric group; it ... 1 The key fact to note is that two elements of the symmetric group are conjugate if and only if they have the same number of k-cycles in their cycle decomposition for all k. In particular, every conjugate of a 3-cycle is a 3-cycle, which solves your second exercise. You have shown in a previous question that$$a(i_1i_2\cdots ...

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hint: $LCD_1 = 5\cdot 4\cdot 3 = 60, LCD_2 = 4\cdot 5 = 20$. Multiply the first equation by $LCD_1 = 60$, and the second by $LCD_2 = 20$. Then simplify both equations each to the form: $ax+by = c$, then finally use elimination to get rid of either variable.

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My book doesn't define what a product of a permutation and a cycle would be. Cycles are permutations! Surely your book told you what it means to compose one permutation with another permutation? Surely your book indicated that cycles are in fact permutations? Please take some time to review the basic concepts before moving forward! Take the cycle ...

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