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 🙂
Best Answer
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 $2, 5$.