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Given multiplicative group of integers modulo 13, namely $\mathbb{Z}_{13}^*$, find all subgroups of this group. I need to prove that this group is cyclic. Also, as $|\mathbb{Z}_{13}^*| = 12$, I know that, if $H$ is a subgroup of $\mathbb{Z}$, then $|H|$ is one of ${1, 2, 3, 4, 6, 12}$. And I got stuck here. No computing.

John Bentin
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Yevs
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  • When you have a finite cyclic group, how many and what are the orders of each subgroup? This have a very definite, precise answer. – Timbuc Dec 12 '14 at 16:07
  • @Timbuc 6 subgroups of orders {1, 2, 3, 4, 6, 12}. But how to find exactly which ones? – Yevs Dec 12 '14 at 16:07
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    @Yves, in fact one single subgroup of each divisor of the group's order. – Timbuc Dec 12 '14 at 16:08
  • You know it must be one of those orders because of Lagrange's Theorem: the order of a subgroup must divide de the order of the group. You now need to identify the elements of those orders, because, for example, if $\circ (a) = 3$, $\vert \langle a \rangle \vert = 3$, because $\langle a \rangle = {0, a, 2a}$ (using additive notation) – Miguelgondu Dec 12 '14 at 16:11

2 Answers2

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If $G=Z^*_{13}$ is cyclic, then every subgroup of $G$ is also cyclic, and so must be generated by a single element. You can find all the subgroups by considering the 12 groups of the form $\langle s \rangle$ for $s\in G$; some of these 12 will be the same, and you need only cross the duplicates off your list of 12.

While considering these 12 subgroups, you will find that one of them, say $\langle i\rangle$, is equal to $G$; this will prove that $G$ is cyclic, and tell you that $G$ is generated by $i$.

MJD
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  • But is it possible not to iterate through them all and find exactly ones I need? – Yevs Dec 12 '14 at 16:13
  • If $a\in\langle b\rangle$, then $\langle a\rangle$ is a subgroup of $\langle b \rangle$. – MJD Dec 12 '14 at 16:27
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that group is the multiplicative group of the field $\mathbb Z_{13}$, the multiplicative group of any finite field is cyclic. For a proof see here.

All you have to do is find a generator (primitive root) and convert the subgroups of $\mathbb Z_{12}$ to those of the group you want by computing the powers of the primitive root.

The subgroups of the group $g,g^2,g^3\dots g^{12}=e$ are those generated by $g^k$ where $k$ divides $12$.

So you want $\langle g \rangle,\langle g^2 \rangle,\langle g^3 \rangle,\langle g^4\rangle, \langle g^6 \rangle, \langle g^{12} \rangle$

Community
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Asinomás
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  • Primitive root is $\overline{2}$. It is equal to $Z_{13}$. Also, $\overline{1}$ is a subgroup. But how to find others? Without iterating them all? – Yevs Dec 12 '14 at 16:10
  • I suggest you first calculate the powers of $2$ and calculate the subgroups I enumerated explicitly, then substitute the powers of $g$ with the numbers you calculated. – Asinomás Dec 12 '14 at 16:15