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        riddles >> putnam exam (pure math) >> Normal subgroups of Q_8 and a 2-group
        
(Message started by: Michael_Dagg on Jan 4^th, 2007, 9:36pm)

Title: Normal subgroups of Q_8 and a 2-group
Post by Michael_Dagg on Jan 4^th, 2007, 9:36pm

Let K be a 2-group. 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 .

Title: Re: Normal subgroups of Q_8 and a 2-group
Post by Eigenray on Jan 25^th, 2007, 6:36am

Suppose x²=1 for all x in K. Then K is abelian. Let H be a subgroup of G=Q₈ 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 Q₈, 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,x²), (-i,x³)}

is not normal, because

(i,x)^(j,1) = (i^j,x¹)=(-i,x)

isn't in H.

Title: Re: Normal subgroups of Q_8 and a 2-group
Post by Icarus on Jan 25^th, 2007, 6:42pm

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 2-group. I thought I knew, but the meaning I had in mind just doesn't make sense with this problem statement.

Title: Re: Normal subgroups of Q_8 and a 2-group
Post by THUDandBLUNDER on Jan 26^th, 2007, 2:45am

on 01/25/07 at 18:42:54, Icarus wrote:

...but what exactly do you mean by a 2-group.

When p is a prime number, then a p-group 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 p-groups by the Sylow theorems, in which case they are called Sylow p-subgroups.

Mathworld (http://mathworld.wolfram.com/p-Group.html)

Title: Re: Normal subgroups of Q_8 and a 2-group
Post by Icarus on Jan 26^th, 2007, 5:31am

Thanks. I was interpreting it as all elements being of order 2, not a power of 2.

Title: Re: Normal subgroups of Q_8 and a 2-group
Post by Michael_Dagg on Mar 2^nd, 2007, 4:49pm

;)

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 2-group and elements of Q_8?

Title: Re: Normal subgroups of Q_8 and a 2-group
Post by Eigenray on Mar 8^th, 2007, 3:48pm

I'm not sure what you're asking, but Q₈ is the smallest non-abelian group all of whose subgroups are normal. In fact, this is true in a very strong sense:

G is a non-abelian group, all of whose subgroups are normal (i.e., a "Hamiltonian group") if and only if G = Q₈ x A, where A is an abelian group with no element of order 4 or infinity.

In particular, the only Hamiltonian p-groups are 2-groups, and these are all of the form discussed in this thread.

The "if" direction is easy: take any (x,y) in Q₈ 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 Q₈, so G = Q₈ x A for some A. If A isn't abelian, then again A contains its own Q₈, hence an element of order 4. And if A contains an element y of order 4 or infinity, then <(i,y)> isn't normal.