Order of elements in a cyclic group ($\mathbb Z_{26}$)

Question

Question: Find the order of $11$ in $\mathbb Z_{26}$.

Answer: $12$

I am having trouble understanding how to find the order of elements of a cyclic (additive) group under this mod $26$. I have a good understanding of how to do this with multiplicative groups ($\mathbb Z_{26}^*$), but having trouble with additive groups.

Please explain how to find the order of elements of $\mathbb Z_{26}$, and in particular how to find the order of $11$.

Answer 1

The order of $11$ is the smallest number $k$ for which $$\underbrace{11 + 11 + \ldots + 11}_{k \textrm{ times}} \equiv 0 \pmod {26}.$$

In other words, we have $11k \equiv 0 \pmod {26}$, which is to say that $11k$ is a multiple of $26$. But $11$ and $26$ are relatively prime, so $k$ itself must be $26$.

Note that $12$ is the order of $11$ in $\mathbb{Z}_{26}^*$, so perhaps there is a typo in the question.