Abstract Algebra- order of a group

**mathatlast** · Jul 1st 2009, 05:13 PM

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

**Chris L T521** · Jul 1st 2009, 05:27 PM

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

In your case,

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

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

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

So two of your answers were correct.

--------------------------------------------------------

As an aside (if you see something like this in the future): If $p$ is a prime, then $\left|U\!\left(p\right)\right|=p-1$

**mathatlast** · Jul 1st 2009, 05:44 PM

Thank you, Chris

**Bruno J.** · Jul 1st 2009, 06:23 PM

And, just to add : more generally,

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

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