Let $G$ be a finite group and suppose it has elements $a, b \in G$ with $o(a) = 10$ and $o(b) = 15$. Let $O(G) = \{m : o(g) = m \text{ for some } g \in G\}$ be the set of all numbers that are orders of elements of G.
(a) Find a subgroup of $G$ having order 3, and show that $3 \in O(G)$.
(b) For each of the following numbers $n$, decide whether or not $O(G)$ must contain $n: n=7, n=5, \text{ and } n=25.$
My attempt:
a) We have $G = \{e, a, a^2,..., a^9, b, b^2,..., b^{14},ab, a^2b,..., ab^2,ab^3,... \}$ and $|G|=1+9+14+9*14=150$. Let $H$ be a subgroup of order 3. By Lagrange theorem $|H|$ divides $|G|$, so if we want to find the subgroup of order 3, we need an element of $G$ of order $3$, hence $H = \{e,b^5,b^{10}\}.$ Since $b^5 \in G$ has order $3, 3 \in O(G)$.
b) To see if $n \in O(G)$, we need to find out if there is any element of G of order n. If there is $n$ must divide |G|. Then since $7\nmid 150$, $5\mid 150$, $25\mid 150$, $n$ can be 5 or 25.
Is this a valid proof? Any corrections would be appreciated.