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December 13th, 2016, 12:18 PM  #1 
Newbie Joined: Dec 2016 From: United States Posts: 3 Thanks: 0  Nonabelian group where G=2p
G is not cyclic and G=2p where p is prime and greater than 2. G is not abelian so we cannot use Cauchy's Theorem directly. a) Show G contains at least one element of order p b) Use a to show that G contains a normal subgroup of order p c) Use b to show that G contains at least one element of order 2 For a, I started with the Lagrange Theorem and stated that x for any x in G has order 2, p or 2p if x is not the identity. How do we show that the order of an element must be p? I'm not sure how to do this because in past problems G was always abelian. Should I still consider a subgroup H: {a, b, ab, e}? Thanks in advance. Last edited by skipjack; December 15th, 2016 at 06:29 AM. 
December 14th, 2016, 01:45 AM  #2 
Senior Member Joined: Feb 2012 Posts: 144 Thanks: 16 
No element can be of order 2p for otherwise G would be cyclic, hence abelian. If all elements (except 1) had order 2, then G would be abelian. To see that, take a,b in G. Then a,b and ab are of order 2, so that aa=bb=abab=1. From abab=1 we get ababb=b i.e. aba=b => abaa=ba => ab=ba 
December 15th, 2016, 05:25 AM  #3 
Senior Member Joined: Sep 2016 From: USA Posts: 167 Thanks: 79 Math Focus: Dynamical systems, analytic function theory, numerics 
This follows from simply counting. You have 2p elements and one is the identity so you have 2p1 nonidentity elements left. If an element has order 2, so does its inverse and thus they occur in pairs which means there is an even number of them. So there must be at least one nonidentity element of order p or 2p. Can you see why it can't have order 2p? 
December 15th, 2016, 11:50 AM  #4 
Senior Member Joined: Feb 2012 Posts: 144 Thanks: 16 
@SDK: an element of order 2 is its own inverse so your argument doesn't work. Besides, in (Z/2Z,+) O is of order 1 and 1 is of order 2 so that there is an odd number of elements of order 2.

January 14th, 2017, 06:29 PM  #5 
Member Joined: Jan 2016 From: Athens, OH Posts: 51 Thanks: 27 
Together with Lagrange, Mehoul has shown there is an element of order p and hence a subgroup H of order p. The index of H in G is two and so (proof?) H is normal in G. For the 3rd question: Let H be a normal subgroup of order p. Then the order of G/H is two and hence cyclic, say G/H is generated by the coset aH. Then 2 divides the order of a. (In any factor group G/H of any finite group G, the order of a coset xH divides the order of x (proof?)). So now if the order of a is 2n, $a^n$ has order 2. 

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