jan31 - Math 4124 Monday January 31 January 31 Ungraded...

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Math 4124 Monday, January 31 January 31, Ungraded Homework Exercise 2.2.1 on page 52 Prove that C G ( A ) = { g G | g - 1 ag = a for all a A } . By definition, g C G ( A ) if and only if gag - 1 = a for all a A . Multiplying on the left by g - 1 and on the right by g , we see that this is if and only if a = g - 1 ag , which yields the required result. Exercise 2.2.2 on page 52 Prove that C G ( Z ( G )) = G and deduce that N G ( Z ( G )) = G . Let g G . Then by definition of center, gx = xg for all x Z ( G ) . Therefore g C G ( Z ( G )) and we have proven that G C G ( Z ( G )) . The reverse inclusion, namely C G ( Z ( G )) G is obvious, and it follows that C G ( Z ( G )) = G , as required. Since the centralizer is always contained in the normalizer, we now see immediately that N G ( Z ( G )) = G . Exercise 2.2.5 on page 52 In each of parts (a) to (c) show that for the specified group G and subgroup A of G , C G ( A ) = A and N G ( A ) = G . (a)
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This note was uploaded on 01/02/2012 for the course MATH 4124 taught by Professor Staff during the Spring '08 term at Virginia Tech.

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jan31 - Math 4124 Monday January 31 January 31 Ungraded...

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