# Value of $\varphi(n)$ and $\lambda(n)$

Is it true that $$\varphi(n)$$ is generally larger than $$\lambda(n)$$ for the same $$n$$? If so, can anyone give me a proof?

As $$n$$ is the product of primes $$p$$ and $$q$$, $$\varphi(n) = (p - 1)(q - 1)$$ and $$\lambda(n) = \operatorname{lcm}(p - 1, q - 1)$$. The former multiplies $$p - 1$$ and $$q - 1$$, whereas the latter finds the least common multiple of the two. Naturally, $$\forall a,b:\operatorname{lcm}(a,b) \le a\cdot b$$, so we can say that $$\forall n:\lambda(n) \le \varphi(n)$$. In the case that the LCM for $$a,b$$ is $$a$$ and $$b$$ themselves, then they will be equal. Otherwise, the LCM will always be smaller. As $$p$$ and $$q$$ are always primes and therefore odd, we have $$2\mid\gcd(p - 1, q - 1)$$, thus $$\lambda(n) < \varphi(n)$$.

So yes, it is true.

• Also note that in the typical case (e.g. RSA) that $p,q$ are both odd, we always have $2\mid\gcd(p-1,q-1)$, hence $\lambda(pq) \mathrel{\color{red}<}\varphi(pq)$. – yyyyyyy May 19 '19 at 8:08
• @yyyyyyy Thanks! Does this edit look correct? – forest May 19 '19 at 8:14