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Could you help me with following excercise?

Find all generators of additive group Z15.

Find all sub-groups of additive group Z15.

Could you please explain how to do that and post a solution?

Thanks, Mark

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closed as off-topic by Jonas Meyer, N. F. Taussig, Alan, Mark Fantini, John Mar 24 at 4:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, N. F. Taussig, Alan, Mark Fantini, John
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In case this is a homework, please add the tag (homework). In any ways, what did you try? In which step did you get stuck? –  user2468 Mar 6 '12 at 17:33
To get started, can you think of one generator for $\mathbf Z_{15}$? Is $2$ a generator? Is $3$? –  Dylan Moreland Mar 6 '12 at 17:35
It might be time consuming, but if you're completely lost try writing out the addition table. –  you Mar 6 '12 at 17:39
Also, what do you know about the size of subgroups as compared to the size of the original group? –  JavaMan Mar 6 '12 at 17:50
The answer is the $10$ (equivalence classes of) numbers from $0$ to $14$ that are relatively prime to $15$. This can be verified painfully, by hand, one at a time. There is of course a shortcut theorem that tells me this, but computing is good. It is easy to verify the others don't work. Start with $1$. Sure. What about $2$? Keep adding $2$ to itself, modulo $15$. For a mild shortcut, note that $2+2+\cdots +2$ (eight of them) is $1$. But since $1$ is a generator, $\dots$. –  André Nicolas Mar 6 '12 at 18:05

2 Answers 2

An element $\overline{a}$ in $\mathbf{Z}_m$ generates if and only if its order is $m$. If $0\leq a\lt m$, then the order of $\overline{a}$ is the least positive integer $k$ such that $m|ka$. Since $a|ka$ for every integer $k$, it follows that the order of $\overline{a}$ is the smallest integer $k$ such that $ka=\mathrm{lcm}(m,a)$.

Under what conditions is $k=m$?

Since $\mathbf{Z}_m$ is cyclic, every subgroup is cyclic. Can you show that if $\overline{a}$ and $\overline{b}$ have the same order, then they generate the same subgroup?

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I'll work with $\mathbf Z_6$. The differences are superficial and translating everything to the situation of $\mathbf Z_{15}$ will be good practice.

For $x \in \mathbf Z$ to be a generator for $\mathbf Z_6$, it is necessary and sufficient that some multiple of $x$ be congruent to $1 \bmod 6$, i.e. that $6 \mid (ax - 1)$ for some integer $a$. To expand this further, there exists an integer $b$ such that $b6 = ax - 1$, so $1 = ax - b6$. What does Bézout now tell you about $a$ and $6$?

It should follow that the classes of $1$ and $5$ are the possible generators.

For finding subgroups, you can do something analogous to how we characterize the subgroups of $\mathbf Z$. In fact, certain theorems make this more than an analogy. If $H$ is a subgroup of $\mathbf Z_6$, then let $y$ be the smallest integer among $\{1, \ldots, 6\}$ whose residue modulo $6$ is in $H$. Again using Bézout, show that $y$ must divide $6$ and that it generates $H$.

You should find that there are four subgroups, generated by the classes of $1$, $2$, $3$, and $6$.

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