How do you multiply n[g]_c? That is, multiply an integer by a number mod c. I am posed with a question in the form of $n[g]_c$. Would I multiply it normally as if $n$ was $[n]_c$? Thanks for any help.
 A: I will answer a somewhat more general question: If $n$ is an integer and $a$ is an element in some arbitrary ring, what does $na$ mean?
There are two ways to view this (they give the same answer):
Multiplying some $x$ by a positive natural number $n$ is the same as adding $x$ to itself $n$ times. Multiplying it by $0$ should of course just give $0$, and multiplying it by a negative integer $n$ would then make the most sense if it meant adding $-x$ to itself $n$ times.
And the above is precisely what we mean when we multiply an element of a ring by an integer. But there is a slightly more abstract way to view this, which explains why the above is the "correct" way to do things.
If $R$ is a ring (with $1$) then there is a unique homomorphism from $\mathbb{Z}$ to $R$, let's call it $\varphi$. So instead of multiplying by some integer $n$, we could multiply by $\varphi(n)$ (which is now an element of $R$, so the multiplication makes sense inside $R$). And indeed, this gives precisely the multiplication I described above. But could we do something differently?
It turns out that no, we cannot, at least as long as we want the multiplication by integers to behave "nicely". By nicely, I mean that we want the usual rules like $(n+m)a = na + ma$ and $(nm)a = n(ma)$, since these rules will mean that the multiplication in fact gives us a homomorphism from $\mathbb{Z}$ to $R$, and hence we are back at the above.
In your specific example, the map $\varphi$ is precisely given by $\varphi(n) = [n]_c$, so your guess was correct.
A: Yeah, Myself is correct when stating that I have not heard of rings. Just started my undergrad abstract algebra class. I am sure we will get to rings eventually though!
