Re: The quaternion group as a product

In article
<14780592.1211173000176.JavaMail.jakarta@xxxxxxxxxxxxx
forum.org>,
Adam Burley <ajburley@xxxxxxxxxxxxxx> wrote:
Is there a known product corresponding to this
situation?

The product of subgroups. If either H or K is
normal
and maximal, then
G = HK provided H and K are distinct, but the
condition is far from
necessary.

Again, PLEASE hit the carriage return when you get to
about 65-70
characters. DO NOT rely on Mathforum's editor to
break up your lines:
it doesn't do it properly. Your reply was one long
line that ran off
the right edge of my screen.


I think it is a problem with your web browser. For example my most recent post at
http://mathforum.org/kb/message.jspa?messageID=6223261
looks fine to me. Perhaps upgrade to the latest version of IE or Firefox?

I am familiar with this product of subgroups, it
being the underlying
idea behind the internal direct/semidirect product.

No, it is not. Because nowhere was there a
requirement that H and K
intersect trivially.


Yes, I was not saying that the example I gave was a direct/semidirect product, only that the product of subgroups is the underlying idea behind those products (with the extra condition that they have trivial intersection and that one/both subgroups be normal).

--
======================================================
================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill
Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

.

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