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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 :)

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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 $2, 5$.

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  • $\begingroup$ Allright, and here we are using the fact that because $\gcd(7,10)=1$ we know that there are $x,y$ such that $7x+10y = 1$ so that $7x=1 \mod 10$ and so some multiple of $(2+5)$ exists that is a generator for the whole set. A similar argument can be made to show that 5 and 7 generate the whole set $\mathbb{Z}$ if I understand correctly? Thanks for your answer :) $\endgroup$ – Slugger Nov 3 '13 at 13:08
  • $\begingroup$ Yes, indeed, you understand correctly. But note that $7 \equiv 7 \pmod{10}, 2\times 7 \equiv 4 \pmod{10}, 3 \times 7 \equiv 1\pmod{10}, 4 \times 7 \equiv 8 \pmod{10} \cdots $. All ten residue classes, are multiples of $7,\,$ $\, \pmod{10}$, i.e. $\langle 7 \rangle = \mathbb Z_{10}$. $\endgroup$ – Namaste Nov 3 '13 at 13:12
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Yes, indeed! Nicely done. ${}$

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  • $\begingroup$ Thanks for your response :) $\endgroup$ – Slugger Nov 3 '13 at 12:55
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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 $G\subseteq S$.

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