# Order of $U(n)$

Let $U(n)$ be group under multiplication modulo $n$. For $n=248$, find number of elements in $U (n)$.

As I tried to do this problem. The number of required elements are $\phi(n)$. So to calculate $\phi(248)$ I first write $248$ as product of powers of primes. So we have $248= 2^3\cdot 31$.

Since $\phi (n) = n (1- \frac{1}{p})(1-\frac{1}{q})$ , where $n= p^iq^j$,

So $\phi (248) =248 (1-\frac{1}{2})(1-\frac{1}{31}) =120$.

But book says answer is $180$. What's going wrong?

• The book is wrong. wolframalpha.com/input/?i=phi%28248%29 Jan 22 '15 at 6:44
• Is the general formula for $\phi (n)$ given in my question correct? Jan 22 '15 at 6:59
• Yes, it is. You can find it here with some other formulas and their proofs. Jan 22 '15 at 7:03
• The even numbers are not invertible $\pmod{248}$ meaning that $\phi(248) \leq 124$. The book is clearly wrong. Oct 11 '20 at 4:01

The euler totient function states that

$$\phi(n) = n \ * \prod_{p|n} (1 - \frac{1}{p})$$

where the product is over the distinct prime numbers dividing n.

Now, by factorisation of 248 into product of primes, we have $$248 = 2^{3} * 31$$.

$$\therefore \ \{2,31\}$$ are the distinct primes dividing 248.

So, $$\phi(248) = 248 \ * (1-\frac{1}{2}) \ * (1-\frac{1}{31})$$

$$= 248 * (\frac{1}{2})*(\frac{30}{31}) = 120$$

"Understanding what a theorem means is a prerequisite to understanding its proof." - Contemporary Abstract Algebra - Gallian, Pg82, Theorem 4.3

The answer indeed is $$120$$ if calculated by the established formulae. Since $$248 =(2^3)\times31$$. Then order of group $$U(248)$$ is $$\varphi \Big(\left(2^3\right)\times31\Big) = \varphi \left(2^3\right) \times \varphi (31) = 2^{3-1} \times (31-1) = 4 \times 30 = 120$$