MATH 113 Study Guide - Midterm Guide: Conjugacy Class, Saw, Normal Subgroup

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12 Oct 2018
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July 16, 2016: (20 points) (a) let g be an element of a group g . Solution: (a) the conjugacy class of g is the orbit of g under the action of g on itself by conjugation. Equivalently, c (g ) = {xgx 1 | x g}. (b) to prove h is normal, we need to show ghg 1 = h for any g . Then every element of ghg 1 has the form ghg 1 for some h h. since h is a union of conjugacy classes, and h h, h c (gi ) for some i. No justi cation required. (a) if f : g h is a homomorphism, and h = g / ker f , then f is surjective. False - the rst permutation is not written as a product of disjoint cycles, so we cannot determine its cycle type without simplifying. Expanding out the product gives (2534)(8713) = (1425387), which has cycle type 7,1.

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