Determine the order of $\frac34+\Bbb Z$ in $\Bbb Q/\Bbb Z$.
The order of an element is the smallest $n$ for which in this case $n(\frac34+\Bbb Z )=e= \Bbb Z$ since the binary operation here is addition.
Now my question is that why isn’t this $n$ being multiplied to $\Bbb Z$, but only to the fraction?
So $1 (\frac34+\Bbb Z ) = (\frac34+\Bbb Z ) $ and $2 (\frac34+\Bbb Z ) =\frac64+\Bbb Z = \frac34 + \Bbb Z$, $3(\frac34+\Bbb Z )= \frac94 + \Bbb Z.$
And lastly $4(\frac34+\Bbb Z )= 3+ \Bbb Z = \Bbb Z$. So $4$ would be the order.
I’m not sure if it’s wrong notation but why isn’t for example $$2 (\frac34+\Bbb Z ) = \frac64+2\Bbb Z$$ instead of $$2 (\frac34+\Bbb Z ) = \frac64+\Bbb Z ?$$
So basically $\Bbb{Q}/\Bbb{Z}$ is a quotient ring, where $\frac{m}{n}\sim\frac{m'}{n'}$, i.e $\frac{m}{n}+\Bbb{Z}=\frac{m'}{n'}+\Bbb{Z}$ iff $\frac{m}{n}=k+\frac{m'}{n'}$ where $k\in\Bbb{Z}$. The way you add (and therefore multiply by an integer) by definition is $(\frac{m}{n}+\Bbb{Z})+(\frac{m'}{n'}+\Bbb{Z})=\frac{m'}{n'}+\frac{m}{n}+\Bbb{Z}$. That's how we define addition - and hence multiplication by an integer.