$C$ groups of highest order Let $G = C_5×C_{12}×C_7$. What is the highest order of element in $G$? How many elements are there in $G$ of the highest order?
We need $\{(a,b,c):a^5=1 \, \, b^{12}=1 \, \, c^7=1 \}$
$(a,b,c)=(1,1,1)^n$
Need the lowest common multiple of $5,12,7$ which is $420$ so $n=420$ right?
For the second part:
$C_5=\{1,x,x^2,...,x^4\}$ 
$1=1$, $(x)^5=1$, $(x^2)^5=1$, and so on so all $5$ elements are of highest order.
For $C_7$, all $7$ elements are of highest order.
But for $C_{12}$, I don't know how to find out which ones are of highest order.
Please help.
Please correct anything I done wrong.

Yes to your answer to the first part: the highest order is equal to the least common multiple of the highest order in each group $C_i$.
You can count the elements of order $420$ in $G$ using the observation that if $a$ is an element of order $5$ in $C_5$, $b$ an element of order $12$ in $C_{12}$ and $c$ and element of order $7$ in $C_7$ then $(a,b,c)$ is an element of highest order in $G$. 
As for $C_{12}$: 
You can think of $C_{12}$ as the group $(\mathbb Z / 13 \mathbb Z)^\times$ with modular multiplication. It should not be too difficult to find all the generators of $(\mathbb Z / 13 \mathbb Z)^\times$.
