I read the above recently. What is this '$p$-component'? What does the double '$\|$' mean? It looks like divide '$|$' but not quite. I understand $p^k | a^t - 1$ but not $p^k\|a^t - 1$
Let $p^k$ be the $p$-component of $a^t - 1$ i.e. $p^k\|a^t - 1$
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$\begingroup$
elementary-number-theory
notation
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2It's from [this answer.](http://math.stackexchange.com/a/156614/242) – 2012-06-17
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The $\|$ means "exactly divides" i.e. the highest power that divides it. For more information read this.
An example would be $5^2\|100$.
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0Ummm, the linked article doesn't actually define "exactly divides" :). – 2012-06-17
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0@ErickWong But it does clarify the notational issue is what I'm hoping. – 2012-06-17
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0@ErickWong: It does. It might seem tedious but one needs to read the article to the end to find out! – 2012-06-17
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0@Gigili: I saw a definition of $\|$, but I didn't see anywhere where it explains what "exactly" means (I fear a naive reader might even misinterpret this as "evenly divides"). – 2012-06-17
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0@ErickWong: You're quite right about. It's about "devide" and just mentioned the name at the end. I'm not sure how it's going to help the reader! (honestly I see no relevance between evenly and exactly) – 2012-06-18
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Another useful way to think about it is that $p^k \parallel a^t-1$ means that $p^k \mid a^t-1$ but $p^{k+1} \nmid a^t-1$.
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0Thank you all for your very clear answers and examples. Does anyone know what "p-component" refers to? – 2012-06-17
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0@Nick: It refers to the Fundamental Theorem of Arithmetic, which states that every $n \in \mathbb N$ can be written uniquely as a product of prime powers $p_1^{r_1} \cdots p_m^{r_m}$. We often think of each integer as being *composed* of these factors, and $p^k \parallel n$ tells you that $p^k$ is one of those components. – 2012-06-18
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0So if a^t - 1 is factored into a product of prime powers – 2012-06-18
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0Now I understand. Thank you! – 2012-06-18
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The notation $p^k \Vert n$ means that the highest power of $p$ dividing the number $n$ is $k$.
Equivalently, we can write $n = p^k \times m$ where $m$ is not divisible by $p$.
For instance, taking $n=20$, we have that $2^2 \Vert 20$ and $5^1 \Vert 20$.