If $t_1\mid p-1$ and $t_2\mid q-1$, then $\text{lcm}(t_1,t_2) \leq \text{lcm}(p-1,q-1)$

Question

Let $p,q $ be primes. It is given that $t_1$ divides $p-1$ and that $t_2$ divides $q-1$. Can it then be proved that $\text{lcm}(t_1,t_2) \leq \text{lcm}(p-1,q-1)$

Answer 1

We don't need primality here

If $a|A, b|B,$ we can prove lcm$(a,b)|$lcm$(A,B)$

For any prime $p,$

if $p^r|a,$ the power of $p$ in $A$ has to be $R\ge r$

and if $p^s|b,$ the power of $p$ in $B$ will be $S\ge s$

So, the power of $p$ in lcm$(a,b)$ will be ${max(r,s)}$

and the power of $p$ in lcm$(A,B)$ will be ${max(R,S)}\ge {max(r,s)}$

So, lcm$(a,b)$ will divide lcm$(A,B)$

Answer 2

By hypothesis $t_1 \mid p-1 , \ t_2 \mid q-1$. So $t_1 \mid [p-1,q-1]$ and $t_2 \mid [p-1,q-1]$. Using the definition of LCM, you can conclude that $[t_1,t_2] \mid [p-1,q-1]$

Note that $[a,b]$ is taken positive by definition so you can conclude your proof.

$[a,b$] is the lcm of a,b

this proof is based on the fact that, if $a \mid b$ then $a \mid kb \ \ \forall k \in \mathbb{Z}$. And on the def. Of lcm: if $m$ is the lcm of $a,b$ then $m$ divides every other multiple of $a$ and $b$.

Answer 3

$t_1 | p-1 \implies p-1 \ge t_1$ and similarly $t_2 \le q-1$

lcm $(t_1,t_2)=kt_1=lt_2$ for some $k,l \in \mathbb{N}$,

Similarly Lcm$(p-1,q-1)=(q-1)w=(p-1)h$ for $k,n \in \mathbb{N}$. Now you can argue that $h \le q-1$ and $w \le p-1$. And thus you have the result.