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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 $p^k\|a^t - 1$

Yai0Phah
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Nick
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3 Answers3

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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 $5^2\|100$.

Yai0Phah
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Eugene
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  • Ummm, the linked article doesn't actually define "exactly divides" :). – Erick Wong Jun 17 '12 at 23:23
  • @ErickWong But it does clarify the notational issue is what I'm hoping. – Eugene Jun 17 '12 at 23:24
  • @ErickWong: It does. It might seem tedious but one needs to read the article to the end to find out! – Gigili Jun 17 '12 at 23:28
  • @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"). – Erick Wong Jun 17 '12 at 23:58
  • @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) – Gigili Jun 18 '12 at 00:15
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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 $p^{k+1} \nmid a^t-1$.

Erick Wong
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  • Thank you all for your very clear answers and examples. Does anyone know what "p-component" refers to? – Nick Jun 17 '12 at 23:54
  • @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. – Erick Wong Jun 18 '12 at 00:03
  • So if a^t - 1 is factored into a product of prime powers – Nick Jun 18 '12 at 00:10
  • Now I understand. Thank you! – Nick Jun 18 '12 at 00:26
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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 $5^1 \Vert 20$.