# Abstract Algebra- order of a group

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• Jul 1st 2009, 05:13 PM
mathatlast
Abstract Algebra- order of a group
Compute |U(4)|, |U(10)|, |U(40)|

I know that I have to compute the order of the group for each. Am I looking for |U(4)|=2, |U(10)|=4 and |U(40)|=10? as the answer?

Thanks in advance
• Jul 1st 2009, 05:27 PM
Chris L T521
Quote:

Originally Posted by mathatlast
Compute |U(4)|, |U(10)|, |U(40)|

I know that I have to compute the order of the group for each. Am I looking for |U(4)|=2, |U(10)|=4 and |U(40)|=10? as the answer?

Thanks in advance

When ever you find the order of $\displaystyle U\!\left(n\right)$ (same as $\displaystyle \mathbb{Z}_n^{\times}$), we are always guaranteed to have 1 and n-1 in $\displaystyle U\!\left(n\right)$, so $\displaystyle \left|U\!\left(n\right)\right|\geq2$ when $\displaystyle n>2$.

In your case,

$\displaystyle U\!\left(4\right)=\{1,3\}$, so $\displaystyle \left|U\!\left(4\right)\right|=2$.

$\displaystyle U\!\left(10\right)=\{1,3,7,9\}$, so $\displaystyle \left|U\!\left(10\right)\right|=4$.

$\displaystyle U\!\left(40\right)=\{1,3,7,9,11,13,17,19,21,23,27, 29,31,33,37,39\}$, so $\displaystyle \left|U\!\left(4\right)\right|=16$.

So two of your answers were correct.

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As an aside (if you see something like this in the future): If $\displaystyle p$ is a prime, then $\displaystyle \left|U\!\left(p\right)\right|=p-1$
• Jul 1st 2009, 05:44 PM
mathatlast
Thank you, Chris
• Jul 1st 2009, 06:23 PM
Bruno J.
And, just to add : more generally,

$\displaystyle U(n)=U(\mathbb{Z}/n\mathbb{Z}) = \phi(n) = n \prod_{p\mid n}\left(1-\frac{1}{p}\right)$