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$\begingroup$

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

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    It's from [this $a$nswer.](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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    @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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    Now 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 $5^1 \Vert 20$.