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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 ?


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My guess is that you're talking about the units of $\mathbf Z/10\mathbf Z$. This is a multiplicative group: if you want to show that $3$ generates it, then you have to look at $3^2 \bmod{10}$, $3^3 \bmod{10}$ and so on. – Dylan Moreland Mar 8 '12 at 16:14
This is what I did , didn't I ? – ron Mar 8 '12 at 16:15
It wasn't so clear to me. If, once you had calculated $6 + 3 = 9$ you noted, "aha, this is $3 \cdot 3 \bmod 10$, now I need to find $3$ times this" then everything is fine. But why the intermediate step of writing down $6$? And once you have that $3^2 = 9$, why not just multiply by $3$ to get $27 \equiv 7$? I don't see any benefit to taking "multiplication is repeated addition" this seriously! – Dylan Moreland Mar 8 '12 at 16:18
Because I won't be able use a calculator in my exam .... thanks :) just wanted to verify . – ron Mar 8 '12 at 16:23
@Dylan - I was confused, and still am, about the point of the question. I thought ron was trying to point out that $3$ is a generator of $U(10)$, but the computation supported a generator for $\mathbb{Z}/10\mathbb{Z}$. Apparently, $3$ generates both the multiplicative group $U(10)$ and the additive group $\mathbb{Z}/10\mathbb{Z}$. Was this what you meant to point out ron? – Chris Leary Mar 8 '12 at 16:54
up vote 3 down vote accepted

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.

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