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Unformatted text preview: S UMMARY OF 11/5 L ECTURE We started by stating explicitly a fact we’ve used several times now: Proposition 1. Suppose N G , and set G = G/N . Then 1. n = 1 if and only if n ∈ N ; and 2. g 1 = g 2 if and only if g 1 g 1 2 ∈ N . Proof. Suppose n = 1 . This is the same as saying that nN = N . Since N is a subgroup, 1 ∈ N whence we deduce that n ∈ N . Suppose instead that n ∈ N . Then nN = N by our favorite homework exercise 2.2(b). But this is the same as saying that n = 1 . This proves the first part of the proposition. Onto the second part: g 1 = g 2 ⇐⇒ ( g 1 )( g 2 ) 1 = 1 ⇐⇒ g 1 g 1 2 = 1 ⇐⇒ g 1 g 1 2 ∈ N. Next we discussed an analogy between integers and groups. What distinguishes a normal subgroup N of G from just any old subgroup H of G is that the quotient G/N is still a group, whereas G/H is not. Thus, N plays a role analogous to that of divisors of an integer: a divisor d of an integer k is special because the quotient k/d is still an integer. We push this analogy further. Recall that any integer can be factored as a product of prime numbers. How does this work? Start with 60, say. The smallest prime factor of 60 is 2; we can thus write 60 = 2 × 30 . The smallest prime factor of 30 is also 2, so we can write 60 = 2 × 2 × 15 . The smallest prime factor of 15 is 3, so 60 = 2 × 2 × 3 × 5 . 1 Since 5 is itself prime, we can no longer find a proper factor, and we’re done. Note that we could have approached this the opposite way: starting with 60, find the largest proper divisor (30), then find the largest proper divisor of that (15), then the largest proper divisor of this (5), and finally the largest proper divisor of this (1). This is the end of the sequence, since 1 has no proper divisors. So, we have the sequence: 1 divides 5 divides 15 divides 30 divides 60 ....
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 Fall '10
 GideonMaschler
 Prime number, Divisor, Cyclic group, largest proper divisor

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