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I need to find all nontrivial subgroups of $G:=\mathbb{Z}_{13}^*$ (with multiplication without zero)

My attempt:

$G$ is cyclic so the order of subgroup of $G$ must be $2,3,4,6$

Now to look for $g\in G$ such that $g^2=e,g^3=e,g^4=e,g^6=e$

$$\begin{align} &12^1=12\mod 13\\ &12^2=1\mod 13\\ \end{align}$$ $\Rightarrow \color{blue}{\{12,1\}}$ is a subgroup of order $2$

$$\begin{align} &3^1=3\mod 13\\ &3^2=9\mod 13\\ &3^3=1\mod 13\\ \end{align}$$

$$\begin{align} &9^1=9\mod 13\\ &9^2=3\mod 13\\ &9^3=1\mod 13\\ \end{align}$$

$\Rightarrow \color{blue}{\{9,3,1\}}$ is a subgroup of order $3$

$$\begin{align} &5^1=5\mod 13\\ &5^2=12\mod 13\\ &5^3=8\mod 13\\ &5^4=1\mod 13\\ \end{align}$$

$$\begin{align} &8^1=8\mod 13\\ &8^2=12\mod 13\\ &8^3=5\mod 13\\ &8^4=1\mod 13\\ \end{align}$$

$\Rightarrow \color{blue}{\{1,5,12,8\}}$ is a subgroup of order $4$

$$\begin{align} &4^1=4\mod 13\\ &4^2=3\mod 13\\ &4^3=12\mod 13\\ &4^4=9\mod 13\\ &4^5=10\mod 13\\ &4^6=1\mod 13\\ \end{align}$$

$$\begin{align} &10^1=10\mod 13\\ &10^2=9\mod 13\\ &10^3=12\mod 13\\ &10^4=3\mod 13\\ &10^5=4\mod 13\\ &10^6=1\mod 13\\ \end{align}$$

$\Rightarrow \color{blue}{\{4,3,12,9,10,1\}}$ is a subgroup of order $6$

Is it correct? is there any easier method?

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  • $\begingroup$ yeah...!!! It is correct $\endgroup$ Commented Jul 7, 2015 at 8:10

2 Answers 2

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Since $2$ is a generator of $\Bbb{Z}_{13}^*$ you have that the subgroups are exactly $$\{ \langle 2^n \rangle : n \mbox{ divides } 12\}$$ i.e. $$\langle 2 \rangle \\ \langle 2^2 \rangle = \langle 4 \rangle \\ \langle 2^3 \rangle = \langle 8 \rangle \\ \langle 2^4 \rangle = \langle 3 \rangle \\ \langle 2^6 \rangle = \langle -1 \rangle \\ \langle 2^{12} \rangle = \langle 1 \rangle \\$$

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  • $\begingroup$ $2^6=12 \mod 13$ $\endgroup$
    – 3SAT
    Commented Jul 7, 2015 at 8:39
  • $\begingroup$ so $1$ is generator? but the order of $\langle 1 \rangle$ is $\infty$ $\endgroup$
    – 3SAT
    Commented Jul 7, 2015 at 8:42
  • $\begingroup$ order of $\langle 1 \rangle=\infty$?? $1$ is identity element $1$ generates trivial subgroup $\endgroup$ Commented Jul 7, 2015 at 8:49
  • $\begingroup$ sorry, the order is one $\endgroup$
    – 3SAT
    Commented Jul 7, 2015 at 8:50
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    $\begingroup$ @Nehorai: $2^6\equiv 12\equiv -1\mod13$. $\endgroup$
    – Bernard
    Commented Jul 7, 2015 at 9:00
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Your solution is right. As, I am noting that(may be i'm wrong), you are applying that "A cyclic subgroup of order $4$ must contain $2$ elements of order $4$ and $1$ element of order $2$, and you searching those elements and listing them. You can reduce your calculation by searching one element of each order, and then you can generate your required subgroups, e.g. $5$ is element of order $4$ so,

$$<5>=\{1,5,8,12\}$$ is subgroup of order $4$

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