# Elements of $\mathbb{Z}_{36}$ that do not generate the whole group.

I am trying to answer this question, and came up with the following elements that do not generate $\mathbb{Z}_{36}$:

$0,2,3,4,6,8,9,10,12,14,15,16,18,20,21,22,24,26,27,28,30$

But the answers don't include some of those elements, for e.g. $8$.

The group generated by $8$ that I've come up with is $\langle{8}\rangle = \{0,4,8,12,16,20,24,28,32\}$, which is the same subgroup generated by $4$.

and for $10$: $\langle{10}\rangle = \{0,2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34\}$, which is the same subgroup generated by $2$.

I cannot figure out what I've done wrong or how elements like $8$ and $10$ generate the whole $\mathbb{Z}_{36}$ group.

• Do you mean that you have some answer sheet which doesn't mention $8$? If yes, it must be an error, because you have the correct subgroups generated by $8$ and $10$. – Arnaud D. Mar 21 '18 at 12:48
• Your answer looks right to me. Either "the answers" are wrong or you have misquoted them or the question in some way. – Ethan Bolker Mar 21 '18 at 12:49
• Your list is missing elements as pointed out by the answer below, and you could have noticed this without any computations: the number of generators of a finite cyclic group is the Euler's totient function of its order. – Pedro Mar 21 '18 at 12:57

Other than that, it is correct. The elements that generate the whole group are precisely those that are coprime with 36, so the list you want is that of those not coprime with 36. As $36=2^2\times3^2$, the list is made precisely by the multiples of 2 and the multiples of 3.
You are looking for the zero divisors, and in $\mathbb{Z}_{36}$ they are the numbers that are not coprime with $36$. This means that the number of zero divisors in your case is $$n-\phi(n)=36-12=\color{orange}{24}$$ If you list them, they are: $$\left\{0,2,3,4,6,8,9,10,12,14,15,16,18,20,21,22,24,26,27,28,30,32,33,34\right\}$$