No simple subgroups in common implies order of groups relatively prime?

Question

Given two finite groups A,B.

If these two groups share no simple subgroups in common, can we conclude that the orders of these two groups are relatively prime?

Answer 1

(1) Groups of prime order are simple.

(2) If a prime $ p$ divides the order of a group, then it has a subgroup of order $p$. (Cauchy's theorem)

Combining the above two answers your question positively. Many times when people say simple group they mean non-abelian simple groups. In such a case take the cyclic group or order $n!$ and the symmetric group $S_n$ (with $n>4$) they have no common non-abelian simple group, however they have the same order.

Answer 2

No : consider $\mathfrak{S}_3=\{Id,r_1,r_2,\tau_1,\tau_2,\tau_3\}.$ You have $o(\tau_1) =o(\tau_2)= 2$ whereas $<\tau_1>=\{Id,\tau_1\}$ and $<\tau_2>=\{Id,\tau_2\}$.