[Math] Proof involving Lagrange’s Theorem

Question

Let $a$ and $b$ be nonidentity elements of different orders in a group $G$ of order $155$.
Prove that the only subgroup of $G$ that contains $a$ and $b$ is $G$ itself.

What I have so far:
We know that by the Lagrange's theorem, $|a|$ and $|b|$ divide $G$. So $|a|$ and $|b|$ must be either $1,5, 31,$ or $155$. Since $a$ and $b$ are non-identity elements, their orders cannot be $1$.

Answer 1

Elementary solution

1. Because the two elements are nonidentity we have : $|a|,|b|>1$
2. Now we have $|a|,|b|\in\{5,31,155\}$ we have two cases:
   • If $|a|$ or $|b|=155$ then the group $|<a,b>|=155$ hence $<a,b>=G$
   • Otherwise $\{|a|,|b|\}=\{5,31\}$ then the group $|<a,b>|=|a||b|=155$ hence $<a,b>=G$