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I'm trying to generate $U_{10}$ (Euler group; multiplicative group of units modulo $10$) with $3$.

If $U_{10}=\{1,3,7,9\}$, let's try to generate all other elements using $3$:

$\begin{align*} \langle 3\rangle = 3 &\longrightarrow 3+3=6\\ &\longrightarrow 6+3=9\\ &\longrightarrow 9+3=12\\ &\longrightarrow 12\bmod 10 = 2\\ &\longrightarrow 2+3 = 5\\ &\longrightarrow 5+3 = 8\\ &\longrightarrow 8+3 =11\\ &\longrightarrow 11\bmod 10 = 1\\ &\longrightarrow 1+3 = 4\\ &\longrightarrow 4+3 = 7\\ &\longrightarrow \text{Done.} \end{align*}$

Is this correct ?

Regards,

  • 0
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1 Answers 1

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No; this is incorrect. You have shown that $3$ generates the additive group of integers modulo $10$. Unfortunately, $U_{10}$ refers to the group of units modulo 10, the multiplicative group modulo $10$.

One reason you should realize that there is something deeply wrong with what you did is that you correctly list the elements of $U_{10}$ as $\{1,3,7,9\}$. However, in the procedure you perform you get other elements! You get $2$, you get $4$, you get $5$, you get $6$, you get $8$, you get $0$,... none of which are supposed to be in your group. Surely, if you were doing things right, you would not be able to get elements not in the group by performing group operations on elements of the group.

You need to consider the multiplicative group; so you want to look at $3$, $3^2$, $3^3$, $3^4$, etc. and see if you get all of $1$, $3$, $7$, and $9$ that way.