MAT-3110 Midterm: MATH 3110 App State Spring2010 Test3 answer key

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15 Feb 2019
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April 20th, 2010: (20 points) for each of the following pairs of groups, if the groups are isomorphic, circle g1 = g2 and explain why they are isomorphic. If the groups aren"t isomorphic, circle g1 6 = g2 and explain why not. (a) c 6 = gl2(r) Both of these are in nite groups (in fact, their cardnality is continuum the size of the real numbers). On the other hand, c is abelian whereas gl2(r) is not: a + b = b + a a, b c (b) z6 = u (7) (cid:20)0 1. Remember that u (n) is the set of all equivalence classes in zn whose representatives are relatively prime to n. since 7 is prime, the only thing we have to throw out is 0. U (7) = {1, 2, 3, 4, 5, 6}. 2 abelian groups of order 6 (both zn and u (m) are always abelian since both modular addition and multiplication are commutative).

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