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Topic: Normal subgroups of Q_8 and a 2group (Read 840 times) 

Michael Dagg
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Normal subgroups of Q_8 and a 2group
« on: Jan 4^{th}, 2007, 9:36pm » 
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Let K be a 2group. Show that every subgroup of Q_8 x K is normal in Q_8 x K iff x^2 equals the identity of K for all x in K .

« Last Edit: Jan 4^{th}, 2007, 9:36pm by Michael Dagg » 
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Eigenray
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Re: Normal subgroups of Q_8 and a 2group
« Reply #1 on: Jan 25^{th}, 2007, 6:36am » 
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Suppose x^{2}=1 for all x in K. Then K is abelian. Let H be a subgroup of G=Q_{8} x K, and let h=(a,x) be an element of H, and let g=(b,y) be an arbitrary element of G. Then h^{g} = (a^{b}, x^{y}) = (a^{b}, x), since K is abelian. But for any two elements a,b of Q_{8}, a direct computation shows that a^{b} = a or a^{1}. In the former case, h^{g}=h, and in the latter, h^{g} = (a^{1},x) = h^{1}, and these are both in H. Conversely suppose K has an element of order >2. Then it has an element x of order 4, and H = <(i,x)> = {(1,1), (i,x), (1,x^{2}), (i,x^{3})} is not normal, because (i,x)^{(j,1)} = (i^{j},x^{1})=(i,x) isn't in H.


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Icarus
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Re: Normal subgroups of Q_8 and a 2group
« Reply #2 on: Jan 25^{th}, 2007, 6:42pm » 
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This problem brings up a question. I'm not sure if my terminology is rusty, or I just never learned this one, but what exactly do you mean by a 2group. I thought I knew, but the meaning I had in mind just doesn't make sense with this problem statement.


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ThudnBlunder
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Re: Normal subgroups of Q_8 and a 2group
« Reply #3 on: Jan 26^{th}, 2007, 2:45am » 
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on Jan 25^{th}, 2007, 6:42pm, Icarus wrote:...but what exactly do you mean by a 2group. 
 When p is a prime number, then a pgroup is a group, all of whose elements have order some power of p. For a finite group, the equivalent definition is that the number of elements in G is a power of p. In fact, every finite group has subgroups which are pgroups by the Sylow theorems, in which case they are called Sylow psubgroups. Mathworld


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Icarus
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Re: Normal subgroups of Q_8 and a 2group
« Reply #4 on: Jan 26^{th}, 2007, 5:31am » 
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Thanks. I was interpreting it as all elements being of order 2, not a power of 2.


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Michael Dagg
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Re: Normal subgroups of Q_8 and a 2group
« Reply #5 on: Mar 2^{nd}, 2007, 4:49pm » 
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Your solution would make a nice (textbook) example. That is saying something because one of keys here is to begin with the most common group that is known, and then etc.... Does anyone have an idea as to why it would be mindful to study the product a of 2group and elements of Q_8?

« Last Edit: Mar 2^{nd}, 2007, 4:53pm by Michael Dagg » 
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Eigenray
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Re: Normal subgroups of Q_8 and a 2group
« Reply #6 on: Mar 8^{th}, 2007, 3:48pm » 
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I'm not sure what you're asking, but Q_{8} is the smallest nonabelian group all of whose subgroups are normal. In fact, this is true in a very strong sense: G is a nonabelian group, all of whose subgroups are normal (i.e., a "Hamiltonian group") if and only if G = Q_{8} x A, where A is an abelian group with no element of order 4 or infinity. In particular, the only Hamiltonian pgroups are 2groups, and these are all of the form discussed in this thread. The "if" direction is easy: take any (x,y) in Q_{8} x A. Since the order of y is finite and not divisible by 4, there's a k such that y^{2+4k}=1, and then any conjugate of (x,y) is either (x,y) itself or (x,y) = (x,y)^{3+4k}. For the "only if" direction, the hard part (especially hard since I don't speak German) is to show that G contains a (complemented) copy of Q_{8}, so G = Q_{8} x A for some A. If A isn't abelian, then again A contains its own Q_{8}, hence an element of order 4. And if A contains an element y of order 4 or infinity, then <(i,y)> isn't normal.


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