h8 - | K | Assume that H is a subgroup of G having order |...

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1 Homework VIII: July 21, 2009 2 . 98 . If G is a group and G/Z ( G ) is cyclic, where Z ( G ) is the center of G , prove that G is abelian. Suppose, on the contrary, that G/Z ( G ) is cyclic, say, with generator aZ ( G ). Since G is not abelian, a / Z ( G ). On the other hand, we now show that a commutes with every g G . Now G is the union of all the cosets of Z ( G ); hence, there is an integer i and an element z Z ( G ) with g = a i z . Therefore, ag = aa i z = a i az = a i za = ga, contradicting a / Z ( G ). 2 . 99 . Let G be a ﬁnite group, let p be prime, and let H be a normal subgroup of G . Prove that if both | H | and | G/H | are powers of p , then | G | is a power of p . If | H | = p h and | G/H | = p q , then | G | = | G/H | · | H | = p h + q . 2 . 105 . Let G be a ﬁnite group with K C G . If ( | K | , [ G : K ]) = 1, prove that K is the unique subgroup of G having order
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Unformatted text preview: | K | . Assume that H is a subgroup of G having order | K | . Since K C G , the quotient group G/K is deﬁned; let π : G → G/K be the natural map (of course, π ( a ) = aK for all a ∈ G ). If h ∈ H , then π ( h ) = hK has ﬁnite order, say, n . By Exercise 2.76(i), we have n | | H | = | K | . But n | | G/K | = [ G : K ], by Lagrange’s Theorem, and so n is a common divisor of | K | and | G/K | . But ( | K | , [ G : K ]) = 1, so that n = 1. Thus, h ∈ ker π = K , and so H ⊆ K . As H and K have the same (ﬁnite) order, weh ave H = K ....
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This note was uploaded on 11/15/2010 for the course MATH 417 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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