Need help with GROUPS.?

Question

I am currently in modern algebra, or also known as contemporary/abstract algebra.

The question is short and simple, but I have absolutely no clue how to approach this problem.

Question: Give an example of a group with 105 elements. Give two examples of groups with 44 elements.

How do I go about this? Can I just name a set and list 105 elements? That seems absurd. I really need help. Any suggestions are greatly appreciated!
Thanks for your time!

Answer 1

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.
No, you can't just name a set with 105 elements
because you don't have a binary operation on it
to make a group.
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!