# Show that λ(m)<Φ(m) for every odd composite number m.

I know that every odd composite number m factors into a set of odd primes. m=p1^(e1)p2^(e2)***pn^(en). I believe that in this situation λ(m)=[(p1-1)^(e1), (p2-1)^(e2), . . ., (pn-n)^(en)], but please correct me if I am wrong. I do not know how to find what Φ(m) is or how it relates to λ(m).

• This is not true. For prime powers $p^k$, we have $\lambda(p^k)=\varphi(p^k)$. – André Nicolas Mar 17 '14 at 22:39
• Could you possibly give a numerical example of this case in which λ(pk)=φ(pk)? – mmm Mar 17 '14 at 22:40
• The smallest composite example is $9$. If there are at least two distinct primes in the factorization, then the result you are looking for is true. – André Nicolas Mar 17 '14 at 22:42
• For 9 we have λ(m)=2 and then we have φ(m) = 6, correct? This means that λ(m)<φ(m). But you have said that there is a case in which λ(m)=φ(m), and that is what I would like an example of if possible. – mmm Mar 17 '14 at 22:45
• $\lambda(9)=6$. I assume that by $\lambda$ you mean the Carmichael function. – André Nicolas Mar 17 '14 at 22:48

Suppose that $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, where the $p_i$ are distinct odd primes. Then $\lambda(n)$ is the least common multiple of the numbers $\varphi(p_1^{a_1}),\dots,\varphi(p_k^{a_k})$.
Also, $\varphi(n)$ is the product of the $\varphi(p_i^{a_i})$, since $\varphi$ is a multiplicative function.
Note that $\varphi(p^a)=p^{a-1}(p-1)$.
If $p$ is odd, then $p-1$ is even, so $\varphi(p^a)$ is even. So if $n$ has two or more distinct odd prime factors, then $2$ is a common divisor of all the $\varphi(p_i^{a_i})$. It follows that the lcm of the $\varphi(p_i^{a_i})$ is less than the product of the $\varphi(p_i^{a_i})$. Thus in that case we have $\lambda(n)\lt \varphi(n)$.
If $n$ is a pure prime power, say $p^a$, then $\lambda(n)=\varphi(n)=p^{a-1}(p-1)$.