The set of elements modulo 35 is a set
Working modulo 35 does not give you a group because we can find zero divisors and those are not invertible.
Multiples of 5 and 7 are zero divisors because they are divisors of 35.
There is no inverse for the elements 5, 7, 10, 14, 15, 20, 21, 25, 28,30 in that set
Listing the elements not on that list gives us a subset consisting of 24 elements as follows
{ 1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34 }
Does the zero divisor 5 generate a group?
Powering of 5 just gives us a series of non-invertible results
{ 5, 25, 20, 30, 10, 15 }
Our generated results do not include the identity element 1 or similar and as we noted there are no inverses.
Does the element 6 generate a group?
We obtain a cyclic group having 2 elements { 6, 1 }
Does the zero divisor 7 generate a group?
Successive powering of 7 just gives us a series of non-invertible results { 7, 14, 28, 21 }
Our generated results do not include the identity element 1 or similar and as we noted there are no inverses.
Does the element 3 generate a multiplicative group?
We obtain a cyclic group having 12 elements
{ 3, 9, 27, 11, 33, 29, 17, 16, 13, 4, 12, 1 }
Does the element 11 generate a multiplicative group?
We obtain a cyclic group having 3 elements { 11, 16, 1 }
Next we attempt to define what a super element is
(i) A sturdy element requires context
[ a set in which it’s properties are special ]
(ii) A sturdy element always generates a group [or stronger] from that enclosing context.
(iii) A sturdy element generates a multiplicative group whose order is
(iv a) A sturdy element shares the context set with an element that generates a maximal group (n-1) elements (depending on the value of our source variable [t])
(iv b) A sturdy element shares the context set with an element that generates a near maximal group (n-1-zero divisors based adjustment) elements (depending on the value of our source variable [t])
For the set
the element -1 + 2^t is a sturdy element
For the set
the element 392+t is not a sturdy element
Setting t=5 we obtain 397 for 392+t
Working modulo 1985 we obtain the following from successive powering of 397
{ 397, 794, 1588, 1191, 397, … }
That subset cannot be a group because it contains a zero divisor in 397
That series of elements is not a group so our element 392+t fails property (ii) and is therefore not a sturdy element.
For the set
the element 1 + 2^t is a sturdy element.
For the set
the element 105+t is not a sturdy element.
Setting t=4 we obtain 109 for 105+t
Working modulo 545 we obtain the following from successive powering of 109 { 109, 436, 109 }
That subset cannot be a group as it contains a zero divisor in 109
That series of elements is not a group so our element 105+t fails property (ii) and is therefore not a sturdy element.
For the set
the element -1 + 3^t is a sturdy element
For the set
the element 38+t is not a sturdy element
Setting t=5 we obtain 43 for 38+t
Working modulo 2107 we obtain the following from successive powering of 43
{ 43, 1849, 1548, 1247, 946, 645, 344, 43, … }
That subset cannot be a group as it contains a zero divisor in 43
That series of elements is not a group so our element 38+t fails property (ii) and is therefore not a sturdy element.
For the set
the element 1 + 3^t is a sturdy element
For the set
the element 25406+t is not a sturdy element
Setting t=5 we obtain 25411 for 25406+t
Working modulo 177877 we obtain the following from successive powering of 25411 { 25411 }
That subset cannot be a group as it contains a zero divisor in 25411
That series of elements is not a group so our element 24506+t fails property (ii) and is therefore not a sturdy element.
Author: Gary Wright 2026
This post appears as the first chapter in a draft book and is available as a pdf