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3 answers

My favorite kind of problem!!
The basic idea is to use the fundamental theorem
of abelian groups.
Since 105 = 3*5*7
we can get a group of 105 elements
by taking Z_3 x Z_5 X Z_7.
The direct product of these 3 cyclic groups gives
a group with 105 elements.
We can also write all this additively
and use the direct sum instead of the direct product.
An actual group with 105 elements is
the group of integers mod 105 under addition.
Now you need 2 groups with 44 elements.
Same idea. Also, I'll give you a way
of getting a concrete example of a group with 44 elements.
Note that 44 = 2²*11
So we can get a group with 44 elements
by taking either Z_4 x Z*_11, the direct
product of cyclic groups of orders 4 and 11
or Z_2 x Z_2 x Z_11, the direct product of a
Klein 4 group with a cyclic group of order 11.
Here is a way to get a concrete example of one
of these groups.
Consider the group U_92, the group of reduced
residues mod 92. The order of U_n is φ(n),
where φ(n) is Euler's φ function, the number
of whole numbers less than n and relatively prime to n.
But φ(92) = φ(4) * φ(23) = 2*22 = 44.
Since this group is abelian(all the U_n are)
I computed the order of every element of U_92 with PARI
and found that every element had order 1, 2, 11 or 22.
So there is no element of order 4 and U_92
must be isomorphic to Z_2 x Z_2 x Z_11.

Whew! Hope that helped!

2007-09-23 05:47:35 · answer #1 · answered by steiner1745 7 · 1 0

Are these Abelian groups, or simple sets? "Abstract algebra" is a pretty big field, so I'm not sure of the restrictions.

If it's Abelian groups, I expect you'll need to factor the numbers and come up with convenient combinations, such as 11 soccer players wearing one of two colored jerseys and being either fresh or tired (granted, 11 different people wont' take on all 44 elements at any one given clock time, but that's not the point of abstract algebra).

If it's simple sets, how about the left and right halves of a piano? 15 weeks? Anything you can enumerate with the requested number of elements. Pick a character set and list 44 characters or 105.

Does that help?

2007-09-22 17:35:37 · answer #2 · answered by norcekri 7 · 0 2

The alkali metals, recent in group a million of the periodic table (earlier frequently going on as group IA), are very reactive metals that don't ensue freely in nature. those metals have purely one electron of their outer shell. consequently, they're waiting to lose that one electron in ionic bonding with different components. as with every metals, the alkali metals are malleable, ductile, and are stable conductors of warmth and electrical energy. The alkali metals are softer than maximum different metals. Cesium and francium are the main reactive components in this group. Alkali metals can explode in the event that they're uncovered to water. An occasion this is Cesium call: Cesium image: Cs Atomic huge type: fifty 5 Atomic Mass: 132.90546 amu form of Neutrons: seventy 8 form of Protons/Electrons: fifty 5 form of capability ranges(Shells) of the Cesium Atom First capability point: 2 2d capability point: 8 third capability point: 18 Fourth capability point: 18 5th capability point: 8 6th capability point: a million The halogens are 5 non-steel components recent in group 17 of the periodic table. The term "halogen" skill "salt-former" and compounds containing halogens are referred to as "salts". All halogens have 7 electrons of their outer shells, giving them an oxidation form of -a million. The halogens exist, at room temperature, in all 3 states of remember: An occasion of Halogens is Chlorine call: Chlorine image: Cl Atomic huge type: 17 Atomic Mass: 35.4527 amu form of Protons/Electrons: 17 form of Neutrons: 18 form of capability ranges(Shells): 3 First capability point: 2 2d capability point: 8 third capability point: 7

2016-12-17 07:56:19 · answer #3 · answered by ? 4 · 0 0

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