Question

Give an example of a group with 105 elements

Answer 1

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.