I have to find $\mathbb{Z}_{p^{3}}(p)$ (p is a prime). I know that the order of any element in $\mathbb{Z}_{p^{3}}$ divides $p^{3}$. How can I then deduce that $\mathbb{Z}_{p^{3}}(p)=\mathbb{Z}_{p^{3}}$?
Finding $\mathbb{Z}_{p^{3}}(p)$
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0What do you mean by saying that every element of $\mathbb{Z}_{p^3}$ has to divide $p^3$? That is not my understanding of what one means by $\mathbb{Z}_{p^3}$. – 2012-04-06
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0It's the cyclic group of order $p^{3}$ and because the order is $p^{3}$ the order of an element in $\mathbb{Z}_{p^{3}}$ has to divide $p^{3}$ – 2012-04-06
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0Your edit has made things clearer. What does $\mathbb{Z}_{p^3}(p)$ mean? – 2012-04-06
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0$\mathbb{Z_{p^{3}}}(p) = \{ u \in Z_{p^{3}} \vert \vert u \vert = p^{n}$ for some $n\geq 0 \}$ – 2012-04-06
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0So isn't this trivial by definition? The only thing that's not contained in the first sentence of your problem is that the divisors of $p^3$ are all of the form $p^n$ for some $n\geq 0$. – 2012-04-06
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0Yes. Why are all divisors of $p^{3}$ of the form $p^{n}$ ? for some $n\geq 0$? – 2012-04-06
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0Ahh, it's true that I assumed $p$ prime. Is that not the case? – 2012-04-06
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0Yes p is a prime – 2012-04-06
2 Answers
So now you have defined $$\Bbb{Z}_{p^3}(p) :=\{ u \in \Bbb{Z}_{p^3}: \operatorname{ord}(u) = p^n \hspace{2mm} \text{for some} \hspace{2mm} n \geq 0\}$$
and are asking why this is equal to $\Bbb{Z}_{p^3}$. It comes down to asking why every element of a group of order $p^3$ must have order $p,p^2$ or $p^3$ right? If you know Lagrange's theorem then the order of any element of the group must be a divisor of $p^3$. So suppose $q$ is any divisor of $p^3$. By the fundamental theorem of arithmetic we can write $q = p_1^{r_1}p_2^{r_2} \ldots p_n^{r_n}$ for distinct primes $p_1, \ldots p_n$ with each $r_i > 0$. Now write out the expression
$$\frac{p}{q}=\frac{p}{p_1^{r_1}p_2^{r_2} \ldots p_n^{r_n}}.$$
We want the right hand side to be an integer. The only way for everything on the bottom to cancel off with the top to give us an integer is when $n=1$. Can you see why this must be the case? If $n > 1$, we have at least one prime factor in the denominator that is different from $p$, say $p_k$ for some $k$ such that $1 \leq k \leq n$. Then $$p_k| p_1^{r_1}p_2^{r_2} \ldots p_n^{r_n} \hspace{2mm} \text{and} \hspace{2mm} p_1^{r_1}p_2^{r_2} \ldots p_n^{r_n}|p$$
which means that $p_k|p$ which is a contradiction because $p_k$ is a prime number different from $p$. So we know now that $q = p_1^{r_1}$, where $r_1 \leq 3$. By the same reasoning as above, $p_1|p$ which forces $p_1 = p$. Hence any divisor $q$ of $p^3$ is of the form $p,p^2$ or $p^3$.
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0By the way if you go on to do ring theory you may want to reconsider using the notation $\Bbb{Z}_{p^3}(p)$ to mean what you meant. In ring theory this means something completely different (people talk about adjoining elements, etc). – 2012-04-06
I think you've tied yourself in a notational knot. Let me read you your question in English:
I have to find all elements whose order is a power of $p$. I know every element has order dividing $p^3$. How can I then deduce that every element has an element which is a power of $p$?
But, of course, this is trivial -- the order of any element is a divisor of $p^3$, the only options for which are $p^0$, $p^1$, $p^2$, and $p^3$, which are all powers of $p$.
Edit: I see I've assumed $p$ prime, without which the result is false. In $\mathbb{Z}_{64}=\mathbb{Z}_{4^3}$, there are elements of order 8, and 8 is not an integral power of $p=4$.
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0How can I see that $p^{0}, p, p^{2}$ and $p^{3}$ are the only options? – 2012-04-06
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0Ahh. Well, this is a part of number theory you should probably work on mastering before talking about groups, subgroups, orders of elements, etc. In any case, this follows from the fact that if $q$ were any other prime factor of $p^n$, we have $q\mid p^n$ implying $q\mid p$, contradiction. – 2012-04-06