Suppose that a group $G$ of order $15$ has only one subgroup of order $3$ and only one subgroup of order $5$, then I need to prove that $G$ is cyclic.
If I can show that $\exists a\in G$ such that $|a|=15$, then the result is proved.
By Lagrange's theorem, every subgroup of order $3$ and $5$ is cyclic. Hence, let $H=\{e,a,a^2\}$ and $K=\{e,b,b^2,b^3,b^4\}$ be the only subgroups of $G$ of orders $3$ and $5$ respectively, such that $|a|=3, |b|=5$ and that $b\notin \langle a \rangle$
Clearly, $H\cup K=\{e,a,a^2,b,b^2,b^3,b^4\}\subset G$. Suppose that $x\in G$ but $x\notin H\cup K$, then by Lagrange's theorem, $|x|=1,3,5,15$. Clearly, $|x|\ne 1$. If $|x|=3$, then $x\in H$ and similarly if $|x|=5$ then $x\in K$. So $|x|=15$. Hence proved.
I can't believe it so I am looking for an example of such a group. How will this group look like? I just can't get past the following thought:
$G=\{e,a,a^2, b,b^2,b^3,b^4, ab, ab^2,ab^3,ab^4,a^2b,a^2b^2,a^2b^3,a^2b^4\}\tag{1}$
Will the group $G$ look like as in $(1)$? If yes, then
$(A)$ how can it be cyclic.
$(B)$ How to ensure closure i.e., whether for example $(ab)^2\in G$ or not?
Thanks for your time.
Well, yes, you can certainly form the group $C_3 \times C_5$ which has the elements that you write above. The point is that... this group is cyclic! And it is generated by the product of the generators.
The reason is that $3$ and $5$ are coprime, and that is actually a necessary condition for this to happen (this = "the direct product of two cyclic groups being cyclic"): in fact, take $C_a = \langle x \rangle$ and $C_b = \langle y \rangle$, where $a$ and $b$ are coprime. Since the generators commute in $G=C_a \times C_b$, the order of $xy$ divides the least common multiple of the orders of $a$ and $b$ (since if $x^a=1$ then $x^{ak}=1$), so if $q$ is a multiple of both $a$ and $b$ then $x^q=y^q=1$ and hence $(xy)^q = 1$ (because, again, they commute).
But if the orders are coprime... the least common multiple is $ab$, the product of the orders! Since $G$ has order $ab$, then it follows that all elements of $G$ are powers of $ab$. In other words, $G$ is cyclic.