When I multiply the set
$$\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$$
by $2$ and take the remainder mod $10$, I get the following repeated pattern.
$$\{2, 4, 6, 8, 0, 2, 4, 6, 8\}$$
Multiplication by any even number creates a similar effect (e.g. multiplication by $4$ is $\{4, 8, 2, 6, 0, 4, 8, 2, 6\}$). But multiplication by odd numbers instead creates a permutation of the elements.
$$ \times 3 = \{3, 6, 9, 2, 5, 8, 1, 4, 7\}\\ \times 9 = \{9, 8, 7, 6, 5, 4, 3, 2, 1\}\\ $$
even when the number is not prime. My question is: why? It clearly has to do with the fact that $2$ and $5$ are the prime factors of $10$, but I'm not sure how. Specifically, I'm wondering:
- Why does multiplication by $k \pmod{n}$ fail to be injective when $\gcd(k, n) \ne 1$?
- Why does multiplying by $2$ split the integers mod $10$ into two identical groups?
- In the case when multiplication by $n$ is injective, is there any way to relate $n$ to the permutation caused by multiplying by it?
I know this is related to group theory, but don't know group theory, so I would appreciate learning the names of the concepts on display here as well.