Show that $\mathbb{Z}_{10}$ is generated by 2 and 5. In the book 'Of Abstract Algebra' by Pinter the following question is asked: 

Show that $\mathbb{Z}_{10}$ is generated by $2$ and $5.\,$ 

Here, $,\mathbb{Z}_{10}\,$ is defined as the group of residues, mod $10$, with the group operation being addition (mod $10$) as usual. 
Now I think I understand why this is true and there are probably many ways to show it. 
First of all I note that:
$$2+2+2+5 \mod 10 \equiv 1$$
Now $1$ is a generator for $\mathbb{Z}_{10}$ because all the elements in $\mathbb{Z}_{10}$ can attained by several repetitions of the group operator on $1.$ Now my question is, is this enough to prove the statement? 
Also, is it sufficiently rigorous or is there a better way to show the desired results? Thanks a lot in advance :)
 A: Yes, indeed, your proof is entirely sufficient, and by showing that $2, 5$ "generate a generator" of the group, you are done. 
You could also simply note that $2 + 5 = 7 = 7\cdot 1$, and since $\gcd(7, 10) = 1$, we know that $7$ generates $\mathbb Z_{10}$. Since $2, 5$ generate $7$, which generates $\mathbb Z_{10}$, it follows that $\mathbb Z_{10}$ is generated by $2, 5$.
A: Yes, indeed! Nicely done. ${}$
A: Yes. In general for many types of structures, if $G$ is a generating set for a structure $S$ (in your example $G=\{1\}$ and $S=\Bbb Z/10\Bbb Z$), then in order to show that another set $F$ generates $S$, it suffices to show that each element of $G$ can be expressed in terms of $F$. The point is one can express any $s\in S$ in terms of elements of $G$, and then substitute their expressions in terms of$~F$ for those elements. For almost any notion of "expression", the substitution produces (something equivalent to) a valid expression.
The condition that each element of $G$ can be expressed in terms of $F$ is also necessary, since $G\subseteq S$.
